Added homeworks from Real Analysis II

Added the first half of the homeworks from Real Analysis II for Spring 2019

Documents/LaTeX/Change of Variables in 2D.tex

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Change of Variables in 2D}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item Evaluate $\displaystyle\int \int_R\ dA$ where $R$ is the region bounded by the lines $x+y=1$, $x+y=2$, $2x-3y=2$, and $2x-3y=5$. Use the transformation $u=x+y$ and $v=2x-3y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. also, compute $x=x(u,v),\ y=y(u,v)$.
+
+		\item Evaluate $\displaystyle\int\int_R \frac{2x-y}{2}\ dA$ where $R$ is the region bounded by the lines $y=2x$, $y=2x-2$, $y=0$, and $y=4$. Use the transformation $u=2x-y$ and $v=y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, computer $x=x(u,v),$ and $y=y(u,v)$.
+		
+		\item Evaluate $\displaystyle\int\int_R 2(x-y)\ dA$ where $R$ is the region bounded by the lines $x=0$, $x=-3$, $y=x$ and $y=x+1$. Use the transformation $u=-x$ and $v=-x+y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$, and $y=y(u,v)$.
+		
+		\item Evaluate $\displaystyle\int\int_R (2x^2-xy-y^2)\ dA$ where $R$ is the region bounded by the lines $y=-2x+4,\ y=-2x+7,\ y=x-2,$ and $y=x+1$ in the first quadrant. Use the transformation $u=x-y$, and $v=2x+y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$ and $y=y(u,v)$.
+		
+		\item Evaluate $\displaystyle\int\int_R (3x^2+14xy+8y^2)\ dA$ where $R$ is the region bounded by the lines $y=-3/2x+1,\ y=-3/2x+3,\ y=-x/4,$ and $y=-x/4+1$. Use the transformation $u=3x+2y$ and $v=x+4y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$, and $y=y(u,v)$.
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 1.tex

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 1}
+\begin{document}
+	\maketitle
+	\textbf{Section 7.1 - The Riemann Integral}
+	\begin{enumerate}
+		\item 
+		\begin{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
+			\item[1.] If $I:=[0,4]$, calculate the norms of the following partitions:
+			\begin{enumerate}
+				\item[c.] $\mathcal{P}_3 := (0,1,1.5,2,3.4,4)$
+				\\$||\mathcal{P}_3||=1.4$
+				\item[d.] $\mathcal{P}_4 := (0,.5,2.5,3.5,4)$
+				\\$||\mathcal{P}_4||=2$
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
+			\item[2.] If $f(x):=x^2$ for $x \in [0,4]$, calculate the following Riemann sums, where $\dotp_i$ has the same partition points as in Exercise 1, and the tags are selected as indicated.
+			\\$\dotp_1 := (0,1,2,4)$
+			\\$\dotp_2 := (0,2,3,4)$
+			\begin{enumerate}
+				\item[(a)] $\dotp_1$ with the tags at the left endpoints of the subintervals.
+				\\The subintervals are:
+				\[I_1:=[0,1],\ I_2:=[1,2],\ I_3:=[2,4]\]
+				So the tags are:
+				\[t_1:=0,\ t_2:=1,\ t_3:=2\]
+				\begin{align*}
+					S(f,\dotp_1) &= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1}) \\
+					&= f(0)(x_1-x_0)+f(1)(x_2-x_1)+f(2)(x_3-x_2) \\
+					&= 0^2(1-0)+1^2(2-1)+2^2(4-2) \\
+					&= 1+8 \\
+					&= 9
+				\end{align*}
+				\item[(b)] $\dotp_1$ with the tags at the right endpoints of the subintervals.
+				\\The subintervals are
+				\[I_1:=[0,1],\ I_2:=[1,2],\ I_3:=[2,4]\]
+				So the tags are:
+				\[t_1:=1,\ t_2:=2,\ t_3:=4\]
+				\begin{align*}
+					S(f,\dotp_1)&=\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1}) \\
+					&= f(1)(x_1-x_0)+f(2)(x_2-x_1)+f(4)(x_3-x_2) \\
+					&= 1^2(1-0)+2^2(2-1)+4^2(4-2)\\
+					&= 1+4+32\\
+					&= 37
+				\end{align*}
+				\item[(c)] $\dotp_2$ with the tags at the left endpoints of the subintervals.
+				\\The subintervals are:
+				\[I_1:=[0,2],\ I_2:=[2,3],\ I_3:=[3,4]\]
+				So the tags are:
+				\[t_1:=0,\ t_2:=2,\ t_3:=3\]
+				\begin{align*}
+					S(f,\dotp_2)&= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1})\\
+					&= f(0)(x_1-x_0)+f(2)(x_2-x_1)+f(3)(x_3-x_2) \\
+					&= 0^2(2-0)+2^2(3-2)+3^2(4-3) \\
+					&= 4+9 \\
+					&= 13
+				\end{align*}
+				\item[(d)] $\dotp_2$ with the tags at the right endpoints of the subintervals.
+				\\So the subintervals are:
+				\[I_1:=[0,2],\ I_2:=[2,3],\ I_3:=[3,4]\]
+				So the tags are:
+				\[t_1:=2,\ t_2:=3,\ t_3:=4\]
+				\begin{align*}
+					S(f,\dotp_2)&= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1})\\
+					&= f(2)(2-0)+f(3)(3-2)+f(4)(4-3) \\
+					&= 2^2(2)+3^2(1)+4^2(1) \\
+					&= 8+9+16 \\
+					&= 33
+				\end{align*}
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[6.]
+			\begin{enumerate}
+				\item[(a)] Let $f(x):=2$ if $0 \leq x < 1$ and $f(x):= 1$ if $1 \leq x \leq 2$. Show that $f \in \mathcal{R}[0,2]$ and evaluate its integral.
+				\\\\We estimate by the graph of $f$ that the integral of $f$ is 3. We must now show by the definition of the integral that the integral of $f$ is 3.
+				\begin{proof}
+					Let $\dotp$ be a tagged partition of $[0,2]$. Let $\dotp_1 \subseteq \dotp$ with tags in $[0,1]$, and let $\dotp_2 \subseteq \dotp$ with tags in $[1,2]$. 
+					\\\\We know that
+					\[[0,1-||\dotp||] \subseteq U_1 \subseteq [0,1+||\dotp||]\ \ \ \ (1)\]
+					and
+					\[[1+||\dotp||,2] \subseteq U_2 \subseteq [1-||\dotp||,2]\ \ \ \ (2)\]
+					where $U_1$ and $U_2$ are the union of the subintervals $\dotp_1$ and $\dotp_2$, respectively.
+					\\\\Now, we can calculate $S(f;\dotp_1)$ and $S(f;\dotp_2)$.
+					\begin{align*}
+						S(f;\dotp_1) &= \sum_{I_i \in \dotp_1} f(t_i)(x_i-x_{i-1}) \\
+						&= \sum_{I_i \in \dotp_1} 2(x_i-x_{i-1}) \\
+						&(I_i \in \dotp_1 \implies I_i \subseteq [0,1]\ \text{where the function value is }2)\\
+						&=2\ \sum_{I_i \in \dotp_1} (x_i-x_{i-1}) \\
+						&\in [2(1-||\dotp||), 2(1+||\dotp||)] = [2-2||\dotp||, 2+2||\dotp||]
+					\end{align*}
+					\begin{center}
+						(Because of (1) we know that the range of the subinterval lengths in $\dotp_1$)
+					\end{center}
+					\begin{align*}
+						S(f;\dotp_2)&=\sum_{I_i \in \dotp_2} f(t_i)(x_i-x_{i-1}) \\
+						&= \sum_{I_i \in \dotp_2} 1 (x_i-x_{i-1}) \\
+						&(I_i \in \dotp_2 \implies I_i \subseteq [1,2]\text{ where the function value is }1) \\
+						&= \sum_{I_i \in \dotp_2} (x_i-x_{i-1}) \\
+						&\in [1-||\dotp||, 1+||\dotp||]
+					\end{align*}
+					\begin{center}
+						(Because of (2), we know the range of the subinterval lengths in $\dotp_2$)
+					\end{center}
+					Therefore,
+						\[S(f;\dotp)=S(f;\dotp_1)+S(f;\dotp_2) \in [3(1-||\dotp||),3(1+||\dotp||)]\]
+						\[\Updownarrow\]
+						\[3-3||\dotp||\leq S(f;\dotp) \leq 3+3||\dotp||\]
+						\[\Updownarrow\]
+						\[|S(f;\dotp)-3| \leq 3||\dotp||\]
+						For arbitrary $\varepsilon >0$ we can pick a tagged partition $\dotp$ such that 
+						\[||\dotp||<\frac{\varepsilon}{3}\]
+						Thus $f \in \mathcal{R}[0,2]$.
+				\end{proof}
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 8 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[8.] If $f \in \mathcal{R}[a,b]$ and $|f(x)| \leq M$ for all $x\in [a,b]$, show that $\abs{\int_{a}^{b}f}\leq M (b-a)$.
+			\\Note that
+			\[-M \leq |f(x)| \leq M,\ \forall\ x \in [a,b]\]
+			By \textit{Theorem 7.1.5 c}, and since every constant function on $[a,b]$ is in $\mathcal{R}[a,b]$, we have that
+			\[-M(b-a) \leq \int_{a}^{b} (-M) \leq \int_{a}^{b} f \leq \int_{a}^{b} M = M(b-a)\]
+			Therefore, 
+			\[\abs{\int_{a}^{b}f}\leq M(b-a)\]
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[12.] Consider the Dirichlet function, introduced in Example 5.1.6(g), defined by $f(x):=1$ for $x \in [0,1]$ rational and $f(x):=0$ for $x \in [0,1]$ irrational. Use the preceding exercise to show that $f$ is \textit{not} Riemann integrable on $[0,1]$.\\
+			\\Let
+			\[\dotp_n:= \left\{\left[\frac{i-1}{n},\frac{i}{n}\right], \frac{i}{n}\right\}_{i=1}^n,\ n \geq 1\]
+			Then $||\dotp_n||=\frac{1}{n} \to 0$ as $n \to \infty$.
+			\\Then,
+			\[S(f;\dotp_n):=\sum_{i=1}^{n}f\left(\frac{i}{n}\right)\left(\frac{i}{n}-\frac{i-1}{n}\right)=\sum_{i=1}^{n}1 \cdot \frac{i}{n} = 1\]
+			because $\frac{i}{n}$ is rational.
+			\\\\Let 
+			\[\dotq_n:=\left\{\left[\frac{i-1}{n},\frac{i}{n}\right],\alpha_i\right\}_{i=1}^n,\ n \geq 1\]
+			where $\alpha_i$ is an irrational number in the interval $\left[\frac{i-1}{n}, \frac{i}{n}\right]$, for $i=1,2,\dots,n$.
+			\\Then $||\dotq_n||=\frac{1}{n} \to 0$ as $n \to \infty$.
+			\\Then,
+			\[S(f;\dotq_n):=\sum_{i=1}^{n}f(\alpha_i)\left(\frac{i}{n}-\frac{i-1}{n}\right)=\sum_{i=1}^{n}0 \cdot \frac{i}{n}=0\]
+			because $\alpha_i$ is irrational.
+			\\\\Therefore,
+			\[\lim\limits_n S(f; \dotp_n)=1 \neq 0 = \lim\limits_n S(f;\dotq_n)\]
+			By the definition of a Riemann integrable function, for any $\varepsilon>0$, there exists $\delta > 0$ such that for all tagged partitions $\dotp$ with $||\dotp||<\delta$ we have
+			\[\abs{S(f;\dotp)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\]
+			Because $||\dotp_n||\to 0$, there exists $n_1 \in \N$ such that
+			\[n>n_1 \implies ||\dotp_n||<\delta\]
+			Similarly, because $||\dotq_n|| \to 0$, there exists $n_2 \in \N$ such that
+			\[n>n_2 \implies ||\dotq_n||<\delta\]
+			Let $n_0:=\max \{n_1,n_2\}$. Then for all $n > n_0$ we have that
+			\[||\dotp_n||<\delta\ \&\ ||\dotq_n||<\delta\]
+			so we have
+			\[\abs{S(f;\dotp_n)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\ \&\ \abs{S(f;\dotq_n)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\]
+			Therefore, for all $n > n_0$,
+			\begin{align*}
+				\abs{S(f;\dotp_n)-S(f;\dotq_n)}&<\abs{S(f;\dotp_n)-\int_{a}^{b}f}+\abs{S(f;\dotq_n)-\int_{a}^{b}f} \\
+				&< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\
+				&= \varepsilon
+			\end{align*}
+			By the definition of the limit of a sequence,
+			\[\lim\limits_n \left[S(f;\dotp_n)-S(f;\dotq_n)\right]=0,\]
+			that is,
+			\[\lim\limits_n S(f;\dotp_n)=\lim\limits_n S(f;\dotq_n)\]
+			which is a contradiction. Therefore $f \notin \mathcal{R}[a,b]$, and hence the Dirichlet function is not Riemann integrable.
+		\end{enumerate}
+	\item 
+		\begin{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 8 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[8.] Suppose that $f$ is continuous on $[a,b]$, that $f(x) \geq 0$ for all $x \in [a,b]$ and that $\int_{a}^{b}f=0$. Prove that $f(x)=0$ for all $x \in [a,b]$.
+			\begin{proof}
+				Suppose there exists $c \in [a,b]$ such that $f(c)>0$. Since $f$ is continuous, there exists $\delta >0$ such that $f(x)>\frac{1}{2}f(c)$ for $x \in (c-\delta, c+ \delta) \subseteq [a,b]$. Then
+				\[\int_{a}^{b}f \geq \int_{c-\delta}^{c+\delta} f \geq \frac{1}{2} f(c) \cdot 2 \delta > 0\]
+				which contradicts the fact that $\int_{a}^{b}f=0$. If $c=a$, then there exists $\delta >0$ such that $f(x)>0$ for $x \in [a,a+\delta)$, and thus the same contradiction is present. The same applies for the case in which $a=b$. Therefore we have that $f(x)=0,\ \forall\ x \in [a,b]$.
+			\end{proof}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 9 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[9.] Show that the continuity hypothesis in the preceding exercise cannot be dropped.
+			\\\\Consider the function $f:[0,1] \to \R$ given by
+			\[f(x):=\begin{cases}
+			1, &x = 0\\
+			0, & x \neq 0
+			\end{cases}\]
+			Then, $f$ has a discontinuity at the point $x=0$ and $\int_{0}^{1} f=0$, but $f$ is not zero everywhere. Therefore, continuity is a necessary part of the hypothesis.\\
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 10 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[10.] If $f$ and $g$ are continuous on $[a,b]$ and if $\int_{a}^{b}f = \int_{a}^{b}g$, prove that there exists $c \in [a,b]$ such that $f(c)=g(c)$.
+			\begin{proof}
+				Let $f$ and $g$ be continuous functions on $[a,b]$ such that
+				\[\int_{a}^{b}f = \int_{a}^{b} g\]
+				Define $h:[a,b] \to \R$ as $h:=f-g$. Then, $h$ is continuous as a difference of continuous functions and
+				\[\int_{a}^{b}h=\int_{a}^{b} (f-g)=\int_{a}^{b}f-\int_{a}^{b}g=0\]
+				Suppose that there exists $c \in [a,b]$ such that $h(c)=0$ since $(f(c)=g(c))$. Then, since $h$ is continuous, it follows that $h(x)>0,\ \forall\ x \in [a,b]$. or $h(x)<0,\ \forall\ x \in [a,b]$ (recall \textit{Bolzano's Theorem}).
+				\\\\Suppose $h(x)>0,\ \forall\ x \in [a,b]$. Then because $h$ is a continuous function on a segment, by the \textit{Maximum-Minimum Theorem} there exists $m>0$ such that
+				\[h(x)\geq m > 0,\ \forall\ x \in [a,b]\]
+				Then we have
+				\[\int_{a}^{b} h \geq \int_{a}^{b} m=m(b-a)>0\]
+				This is a contradiction with the fact that $\int_{a}^{b} h=0$.
+				\\\\Now, for the case in which $h(x)<0,\ \forall\ x \in [a,b]$, by the \textit{Maximum-Minimum Theorem} we know that there exists $M<0$ such that
+				\[h(x)\leq M < 0\ \forall\ x \in [a,b]\]
+				and thus
+				\[\int_{a}^{b} h \leq \int_{a}^{b} M \leq M(b-a)<0\]
+				which again yields a contradiction.
+				\\\\Therefore, there exists $c \in [a,b]$ such that $h(c)=0$, that is, $f(c)=g(c)$.
+			\end{proof}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 13 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[13.] Give an example of a function $f:[a,b] \to \R$ that is in $\mathcal{R}[c,b]$ for every $c \in (a,b)$ but which is not in $\mathcal{R}[a,b]$.
+			\\\\Define a function $f$ on $[0,1]$ by
+			\[f(x):=\begin{cases}
+			\frac{1}{x}, &x \in (0,1] \\
+			1, &x=0
+			\end{cases}\]
+			For every $c>0,\ f \in \mathcal{R}[c,1]$ because $f$ is continuous on $[c,1]$.
+			\\\\Now, let's show that $f$ isn't Riemann integrable on $[0,1]$.
+			\\\\Define a tagged partition to be
+			\[\dotp := \left\{\left[\frac{i-1}{n},\frac{i}{n}\right],\frac{i}{n}\right\}_{i=1}^n\]
+			Then
+			\begin{align*}
+				S(f;\dotp) &= \sum_{i=1}^{n} f \left(\frac{i}{n}\right)\left(\frac{i}{n}-\frac{i-1}{n}\right) \\
+				&= \sum_{i=1}^{n}\frac{1}{\frac{i}{n}} \cdot \frac{1}{n} \\
+				&= \sum_{i=1}^{n} \frac{1}{i}
+			\end{align*}
+			As $n \to \infty$, $S(f;\dotp)$ diverges (since it is a harmonic series). Thus, $f$ is not Riemann integrable on $[0,1]$.
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Use the right-endpoint Riemann sums to evaluate the following integrals:
+		\begin{enumerate}
+			\item $\int_{2}^{5}(3x-1)dx$
+			\begin{align*}
+				\lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x_i &= \sum_{i=1}^{n} f(a+i\Delta x) \Delta x \\
+				&= \sum_{i=1}^{n} f\left(2+i\left(\frac{3}{n}\right)\right)\cdot \left(\frac{3}{n}\right) \\
+				&= \sum_{i=1}^{n} \left(3 \cdot \left(2+\frac{3i}{n}\right)-1\right) \cdot \left(\frac{3}{n}\right) \\
+				&= \sum_{i=1}^{n} \left(6+\frac{9i}{n}-1\right)\left(\frac{3}{n}\right) \\
+				&= \sum_{i=1}^{n}\left(5+\frac{9i}{n}\right)\left(\frac{3}{n}\right) \\
+				&= \sum_{i=1}^{n} \frac{15}{n} +\frac{27i}{n^2} \\
+				&= \frac{15}{n} \sum_{i=1}^{n} 1 + \frac{27}{n^2} \sum_{i=1}^{n} i \\
+				&= \frac{15}{\cancel{n}} \cdot \cancel{n} + \frac{27}{n^2} \cdot \frac{n(n+1)}{2} \\
+				&= 15 + \frac{27}{n^2} \cdot \frac{n^2+n}{2} \\
+				&= 15 + \frac{27n^2+27n}{2n^2} \\
+				\int_{2}^{5} (3x-1)dx &= \lim\limits_{n \to \infty} 15+\frac{27n^2+27n}{2n^2}\\
+				&= 15+\frac{27}{2}\\
+				&=28.5
+			\end{align*}
+			\item $\int_{0}^{4} (x^2+2x)dx$
+			\begin{align*}
+				\int_{0}^{4} (x^2+2x)dx &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(a+i\Delta x\right) \Delta x \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(0+i\left(\frac{4}{n}\right)\right)\cdot \left(\frac{4}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left[\left(\frac{4i}{n}\right)^2+2\left(\frac{4i}{n}\right) \cdot \right]\left(\frac{4}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{16i^2}{n^2}+\frac{8i}{n}\right)\cdot \left(\frac{4}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{64i^2}{n^3}+\frac{32i}{n^2}\right) \\
+				&= \lim\limits_{n \to \infty} \left(\frac{64}{n^3}\ \sum_{i=1}^{n} i^2 + \frac{32}{n^2}\ \sum_{i=1}^{n} i\right) \\
+				&= \lim\limits_{n \to \infty} \left(\frac{64}{n^3}\cdot\frac{n(n+1)(n+2)}{6}+\frac{32}{n^2}\cdot \frac{n(n+1)}{2}\right) \\
+				&= \lim\limits_{n \to \infty} \left(\frac{64n^3+192n^2+128n}{6n^3}+\frac{32n^2+32n}{2n^2}\right) \\
+				&= \frac{64}{6} + 16 \\
+				&\approx 26.6667
+			\end{align*}
+			\item $\int_{0}^{2} (2x^3+x)dx$
+			\begin{align*}
+				\int_{0}^{2} (2x^3+x)dx &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(a+i\Delta x)\Delta x \\
+				&=\lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(\frac{2i}{n}\right)\left(\frac{2}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(2\left(\frac{2i}{n}\right)^3+\frac{2i}{n}\right)\left(\frac{2}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{32i^3}{n^4}+\frac{4i}{n^2}\right) \\
+				&= \lim\limits_{n \to \infty} \left[\frac{32}{n^4}\ \sum_{i=1}^{n}i^3+\frac{4}{n^2}\ \sum_{i=1}^{n}i\right] \\
+				&= \lim\limits_{n \to \infty} \left[\frac{32}{n^4}\cdot\frac{n^2(n+1)^2}{4}+\frac{4}{n^2}\cdot\frac{n(n+1)}{2}\right] \\
+				&= \lim\limits_{n \to \infty} \left[\frac{32n^4+64n^3+32n^2}{4n^4}+\frac{4n^2+4n}{2n^2}\right] \\
+				&= 8+2 \\
+				&= 10
+			\end{align*}
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Express each of the following as a definite integral. Then use calculus to evaluate the integral.
+		\begin{enumerate}
+			\item $\lim\limits_{|P| \to 0}\ \sum\limits_{i=1}^{n} \left(\frac{3}{w_i^2}\right) \Delta x_i$ where $P$ is a partition of $[1,3]$.
+			\begin{align*}
+				\lim\limits_{|P| \to 0}\ \sum\limits_{i=1}^{n} \left(\frac{3}{w_i^2}\right) \Delta x_i &= \int_{1}^{3} x dx \\
+				&= \left.\frac{x^2}{2} \right|_1^3 \\
+				&= \frac{3^2}{2} - \frac{1^2}{2} \\
+				&= \frac{9}{2} - \frac{1}{2} \\
+				&= \frac{8}{2} \\
+				&= 4
+			\end{align*}
+			\item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(3 + \frac{2i}{n}\right)^2\cdot \left(\frac{2}{n}\right)$
+			\begin{align*}
+				\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(3 + \frac{2i}{n}\right)^2\cdot \left(\frac{2}{n}\right) &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\Delta x\right)^2\cdot \left(\Delta x\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)^2 \cdot \left(\frac{b-a}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(3+i\left(\frac{2}{n}\right)\right)^2 \cdot \left(\frac{2}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(3+i\left(\frac{5-3}{n}\right)\right)^2\cdot \left(\frac{5-3}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\cdot \Delta x \\
+				&= \int_{3}^{5} f(t_i) dx \\
+				&= \int_{3}^{5} x^2 dx \\
+				&= \left. \frac{x^3}{3} \right|_{3}^{5} \\
+				&= \frac{5^3}{3}-\frac{3^3}{3} \\
+				&= \frac{125}{3} - \frac{27}{3} \\
+				&= \frac{125}{3} - 9 \\
+				&\approx 32.66667
+			\end{align*}
+			\item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(1 + \frac{4(i-1)}{n}\right)^5\cdot \left(\frac{4}{n}\right)$
+			\begin{align*}
+				\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(1 + \frac{4(i-1)}{n}\right)^5\cdot \left(\frac{4}{n}\right) &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\Delta x\right)^5\cdot \left(\Delta x\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)^5 \cdot \left(\frac{b-a}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(1+\frac{4i-4}{n}\right)^5\cdot\left(\frac{4}{n}\right)\\
+				&=\int_{1}^{5} f(t_i)dx \\
+				&= \int_{1}^{5} x^5 dx \\
+				&= \left. \frac{x^6}{6} \right|_1^5 \\
+				&= \frac{5^6}{6}-\frac{1^6}{6} \\
+				&= \frac{15625}{6}-\frac{1}{6} \\
+				&= \frac{15624}{6} \\
+				&= \frac{7812}{3} \\
+				&= 2604
+			\end{align*}
+			\item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{i^2}{n^3}$
+			\begin{align*}
+				\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{i^2}{n^3} &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} (a+i\Delta x)\Delta x \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)\cdot\left(\frac{b-a}{n}\right) \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(0+i\left(\frac{1-0}{n}\right)\right)^2 \cdot \left(\frac{1-0}{n}\right) \\
+				&= \int_{0}^{1} f(t_i)dx \\
+				&= \int_{0}^{1} x^2 dx \\
+				&= \left.\frac{x^3}{3}\right|_0^1 \\
+				&= \frac{1^3}{3} - \frac{0^3}{3} \\
+				&= \frac{1}{3}
+			\end{align*}
+			\item Show that $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{n}{n^2+i^2}=\frac{\pi}{4}$
+			\begin{align*}
+				\lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{n}{n^2+i^2} &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{n}{n^2+i^2}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \\
+				&= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(\frac{i}{n})^2} \cdot \frac{1}{n} \\
+				&= \int_{0}^{1} \frac{1}{x^2} dx \\
+				&= \arctan(x) |_0^1 \\
+				&= \arctan(1)-\arctan(0) \\
+				&= \frac{\pi}{4} - 0 \\
+				&= \frac{\pi}{4}
+			\end{align*}
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Give examples of functions $f:[0,1] \to \R$ such that
+		\begin{enumerate}
+			\item $f \notin \mathcal{R}[0,1]$, but $|f|$ and $f^2$ are both in $\mathcal{R}[0,1]$.
+			\\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			1 &x \in \Q \\
+			-1 &x \in \R\setminus\Q
+			\end{cases}$, and let $P:=\{x_0,x_1,\dots, x_n\}$ be any partition of $[0,1]$. Then $M_i=1$ and $m_i=-1$ for all $i=1,2,\dots,n$. thus $U(f,P)=1$ and $L(f,P)=-1$ for all $P$. Thus $U(f)=1$ and $L(f)=-1$. Thus $f$ is not integrable.
+			\\\\However, $|f|(x)=1$ for all $x \in [0,1]$. Since $|f|$ is a continuous function $|f|$ is integrable on $[0,1]$, and also since $f^2(x)=1$ is also a continuous function, we have that $f^2$ is also integrable on $[0,1]$.\\
+			\item $f$ is bounded, but $f \notin \mathcal{R}[0,1]$.
+			\\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			1 &x \in \Q \\
+			0 &x \in \R\setminus\Q
+			\end{cases}$. Let $P$ be any partition of $[0,1]$, then
+			\[U(P;f):=\sum M_i \Delta x_i = \sum 1 \Delta x_i = b-a\]
+			and
+			\[L(P;f)=\sum m_i \Delta x_i = \sum 0 \Delta x_i = 0(b-a)=0\]
+			Hence
+			\[\overline{\int_{0}^{1}} fdx=b-a \neq 0 = \underline{\int_{0}^{1}} f dx \]
+			Hence $f$ is not Riemann integrable.
+			\item $f \in \mathcal{R}[0,1]$ and $f$ is not monotone.
+			\\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			3 & 0 \leq x < \frac{1}{3} \\
+			1 & \frac{1}{3} \leq x < \frac{2}{3} \\
+			3 & \frac{2}{3} \leq x \leq 1
+			\end{cases}$. The function $f$ is non-monotonic on $[0,1]$.
+			\\\\The upper Riemann integral of $f$ is
+			\[\overline{\int_{[0,1]}}f=\inf \left\{\int_{[0,1]} g: g \text{ is a piecewise constant on }[0,1] \text{ and } g(x) \geq f(x)\ \forall\ x \in [0,1]\right\}\]
+			\[=\frac{7}{3}\]
+			Similarly, the lower integral of $f$ is given by
+			\[\underline{\int_{[0,1]}}f = \sup \left\{\int_{[0,1]} g: g \text{ is a piecewise constant on } [0,1] \text{ and } g(x) \leq f(x)\ \forall\ x \in [0,1]\right\}\]
+			\[=\frac{7}{3}\]
+			Since $\displaystyle\overline{\int_{[0,1]}}f=\underline{\int_{[0,1]}} f$, the function $f$ is Riemann integrable on $[0,1]$ and $\displaystyle\int_{[0,1]} f=\frac{7}{3}$.
+			\item $f \in \mathcal{R}[0,1]$ and $f$ is neither monotone nor continuous.
+			\\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			0 & x \in \{0,1\}\cup([0,1]\setminus\Q) \\
+			\frac{1}{q} & x \in (0,1) \cap \Q,\ x=\frac{p}{q},\ p,q \in \N, \text{ and}\\
+			& \text{$p,q$ are relatively prime}
+			\end{cases}$. We note that $f$ is known as the Riemann function. Thus it is well known that this function is not piecewise continuous nor is it monotone.
+		\end{enumerate}	
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Prove or justify, if true or provide a counterexample, if false.
+		\begin{enumerate}
+			\item If $f(x) \leq g(x) \leq h(x)$ for all $x \in [a,b]$, and $f,h \in \mathcal{R}[a,b]$, then so is $g \in \mathcal{R}[a,b]$.\\\\
+			This is true by the \textit{Squeeze Theorem}.\\
+			\item If $f \in \mathcal{R}[a,b]$, then $f$ is continuous on $[a,b]$.
+			\\\\This is a false statement. Consider $f:[0,3] \to \R$ given by $f(x):=\begin{cases}
+			2,\ &0 \leq x \leq 1 \\
+			3,\ &1 < x \leq 3
+			\end{cases}$
+			Then we have that $\int_{0}^{3} f(x)=8$, and thus $f \in \mathcal{R}[0,3]$, but $f$ is not continuous.\\
+			\item If $|f| \in \mathcal{R}[a,b]$, then $f \in \mathcal{R}[a,b]$.
+			\\\\This is a false statement. Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			1 &x \in \Q \\
+			-1 &x \in \R\setminus\Q
+			\end{cases}$, and let $P:=\{x_0,x_1,\dots, x_n\}$ be any partition of $[0,1]$. Then $M_i=1$ and $m_i=-1$ for all $i=1,2,\dots,n$. thus $U(f,P)=1$ and $L(f,P)=-1$ for all $P$. Thus $U(f)=1$ and $L(f)=-1$. Thus $f$ is not integrable.
+			\\\\However, $|f|(x)=1$ for all $x \in [0,1]$. Since $|f|$ is a continuous function $|f|$ is integrable on $[0,1]$.\\
+			\item Let $f$ be bounded on $[a,b]$. If $P$ and $Q$ are partitions of $[a,b]$, then $P \cup Q$ is a refinement of both $P$ and $Q$.
+			\\\\This is a true statement because this satisfies the definition of a refinement, since both $P \subseteq P \cup Q$ and $Q \subseteq P \cup Q$.\\
+			\item If $f$ is continuous on $[a,b)$ and on $[b,c]$, then $f \in \mathcal{R}[a,c]$.
+			\\\\This is a false statement. Consider the function $f:[0,5] \to \R$ given by $f(x):= \begin{cases}
+			\frac{x}{x-2}, &0 \leq x < 2 \\
+			0, & 2 \leq x \leq 5
+			\end{cases}$. Since an asymptote exists and is unbounded on $[0,5]$, we have that $f$ is not Riemann integrable.
+			\item If $f,g\in\mathcal{R}[a,b]$, then $f-g \in \mathcal{R}[a,b]$.
+			\\\\This is true by \textit{Theorem 7.1.5 c}, since it can be rewritten as $f+(-g)$.\\
+			\item If $f$ is monotone on $[a,b]$, then $f \in \mathcal{R}[a,b]$.
+			\\\\This is a true statement by \textit{Theorem 7.2.6}:
+			\begin{theorem*}
+				If $f:[a,b]\to\R$ is monotone on $[a,b]$, then $f \in \mathcal{R}[a,b]$.
+			\end{theorem*}
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 2.tex

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 2}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item \textbf{Section 7.4}
+		\begin{enumerate}
+			\item[1.] Let $f(x):=|x|$ for $-1 \leq x \leq 2$. Calculate $L(f;P)$ and $U(f,P)$ for the following partitions:
+			\begin{enumerate}
+				\item[(a)] $\mathcal{P}_1 :=(-1,0,1,2)$
+				\\\\Our terms are:
+				\[x_0:= -1,\ \ x_1:=0,\ \ x_2:=1,\ \ x_3:=2\]
+				and our intervals are:
+				\[I_1:=[-1,0],\ \ I_2:=[0,1],\ \ I_3:=[1,2]\]
+				thus $L(f,\mathcal{P}_1)$ is:
+				\begin{align*}
+					L(f,\mathcal{P}_1) &= \sum_{i=1}^{n} m_i(x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} \left(\inf\{f(x):x \in [x_{i-1},x_i]\}\right) (x_i-x_{i-1}) \\
+					&= (\inf \{|x|:x \in [-1,0]\})(0-(-1))\\ 
+					&+ (\inf \{|x|: x \in [0,1]\})(1-0)\\
+					&+(\inf \{|x|: x\in [1,2]\})(2-1) \\
+					&= 0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1 \\
+					&= 0+0+1 \\
+					&= 1
+				\end{align*}
+				and
+				\begin{align*}
+					U(f,\mathcal{P}_1) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} sup \{f(x): x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} \sup \{|x|: x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\
+					&= (\sup \{|x|:x \in [-1,0]\})(0-(-1)) \\
+					&+ (\sup \{|x| : x \in [0,1]\})(1-0) \\
+					&+ (\sup \{|x|: x \in [1,2]\})(2-1) \\
+					&= 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 \\
+					&= 1 + 1 +2 \\
+					&= 4
+				\end{align*}
+				So, $L(f,\mathcal{P}_1)=1$ and $U(f,\mathcal{P}_1)=4$\\
+				\item[(b)] $\mathcal{P}_2 := (-1,-1/2,0,1/2,1,3/2,2)$.
+				\\\\Our terms are:
+				\[x_0:=-1,\ \ x_1:=-\frac{1}{2},\ \ x_2:=0,\ \ x_3:=\frac{1}{2},\ \ x_4:= 1,\ \ x_5:=\frac{3}{2},\ \ x_6:=2\]
+				and our intervals are:
+				\[I_1:=\left[-1,-\frac{1}{2}\right],\ \ I_2:=\left[-\frac{1}{2},0\right],\ \ I_3:=\left[0,\frac{1}{2}\right],\ \ I_4:=\left[\frac{1}{2},1\right],\ \ I_5:=\left[1,\frac{3}{2}\right],\ \ I_6:=\left[\frac{3}{2}, 2\right]\]
+				So $L(f,\mathcal{P}_2)$ is
+				\begin{align*}
+					L(f,\mathcal{P}_2) &= \sum_{i=1}^{n} m_i(x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} \inf \{f(x): x \in [x_{i-1}, x_i]\}(x_i-x_{i-1}) \\
+					&= \inf\left\{|x|: x \in \left[-1,-\frac{1}{2}\right]\right\}\left(-\frac{1}{2}-(-1)\right) \\
+					&+ \inf\left\{|x|: x \in \left[-\frac{1}{2}, 0\right]\right\} \left(0-\left(-\frac{1}{2}\right)\right) \\
+					&+ \inf\left\{|x|: x \in \left[0,\frac{1}{2}\right]\right\}\left(\frac{1}{2}-0\right) \\
+					&+ \inf \left\{|x|: x \in \left[\frac{1}{2}, 1\right]\right\}\left(1-\frac{1}{2}\right) \\
+					&+ \inf\left\{|x|: x \in \left[1,\frac{3}{2}\right]\right\}\left(\frac{3}{2}-1\right) \\
+					&+ \inf \left\{|x|: x \in \left[\frac{3}{2}, 2\right]\right\}\left(2 - \frac{3}{2}\right) \\
+					&= \frac{1}{2} \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{2} \\
+					&=\frac{1}{4} + 0 + 0+ \frac{1}{4} + \frac{1}{2} + \frac{3}{4} \\
+					&= \frac{7}{4}
+				\end{align*}
+				and
+				\begin{align*}
+					U(f,\mathcal{P}_2) &= \sum_{i=1}^{n} M_i(x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} \sup \{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+					&= \sup \left\{|x|: x \in \left[-1,-\frac{1}{2}\right]\right\}\left(-\frac{1}{2}-(-1)\right) \\
+					&+ \sup\left\{|x|: x \in \left[-\frac{1}{2},0\right]\right\}\left(0-\left(-\frac{1}{2}\right)\right) \\
+					&+ \sup\left\{|x|: x \in \left[0,\frac{1}{2}\right]\right\}\left(\frac{1}{2}-0\right) \\
+					&+ \sup\left\{|x|: x \in \left[\frac{1}{2}, 1\right]\right\}\left(1-\frac{1}{2}\right) \\
+					&+ \sup\left\{|x|: x \in \left[1, \frac{3}{2}\right]\right\}\left(\frac{3}{2}-1\right) \\
+					&+ \sup\left\{|x|: x \in \left[\frac{3}{2}, 2\right]\right\}\left(2-\frac{3}{2}\right) \\
+					&= 1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} \\
+					&= \frac{1}{2} + \frac{1}{4} + \frac{1}{4}+ \frac{1}{2} + \frac{3}{4} + 1 \\
+					&= \frac{13}{4}
+				\end{align*}
+				So $L(f,\mathcal{P}_2) = \frac{7}{4}$ and $U(f,\mathcal{P}_2) = \frac{13}{4}$\\
+			\end{enumerate}
+			\item[2.] Prove if $f(x):=c$ for $x \in [a,b]$, then its Darboux integral is equal to $c(b-a)$.
+			\begin{proof}
+				Let $\mathcal{P} := (x_0, x_1, \dots, x_n)$ be a partition of $[a,b]$ where
+				\[a=x_0 < x_1 < x_2 < \dots < x_n=b\]
+				then $M_i:= \sup f(x)=c$ since $f$ is constant, for all $x \in [x_{i-1}, x_i]$, and $m_i:= \inf f(x)=c$ again since $f$ is constant, for all $x \in [x_{i-1},x_i]$.
+				\\\\Then we have that
+				\begin{align*}
+					U(f, \mathcal{P}) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} c (x_i-x_{i-1}) \\
+					&= c \sum_{i=1}^{n} (x_i-x_{i-1}) \\
+					&= c (x_n-x_0) \\
+					&= c (b-a), &\text{as both $b$ and $a$ were defined for $\mathcal{P}$}
+				\end{align*}
+				So $U(f,\mathcal{P}):= c(b-a)$. As for $L(f,\mathcal{P})$:
+				\begin{align*}
+					L(f,\mathcal{P}) &= \sum_{i=1}^{n} m_i (x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} c (x_i-x_{i-1}) \\
+					&= c \sum_{i=1}^{n} (x_i-x_{i-1}) \\
+					&= c (x_n-x_0) \\
+					&= c(b-a), &\text{as both $b$ and $a$ were defined for $\mathcal{P}$}
+				\end{align*}
+				and thus $L(f,\mathcal{P}) = c(b-a)$.
+				\\\\Now we must find the Darboux integral of $f(x)$. So we have that the upper Darboux integral of $f(x)$ is 
+				\begin{align*}
+					U(f) &= \inf \{U(f,\mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\
+					&= \inf \{c(b-a): \mathcal{P} \in \mathscr{P}[a,b]\} \\
+					&= c(b-a)
+				\end{align*}
+				and the lower Darboux integral is
+				\begin{align*}
+					L(f) &= \sup \{L(f, \mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\
+					&= \sup \{c(b-a) : \mathcal{P} \in \mathscr{P}[a,b]\} \\
+					&= c(b-a)
+				\end{align*}
+				Thus we have that $U(f)=L(f)$, which yields that $f$ is Darboux integrable on $[a,b]$ and the Darboux integral of $f$ is $c(b-a)$.
+			\end{proof}
+			\item[3.] Let $f$ and $g$ be bounded functions on $I:=[a,b]$. If $f(x) \leq g(x)$ for all $x \in I$, show that $L(f) \leq L(g)$ and $U(f) \leq U(g)$.
+			\begin{proof}
+				Let $f,g$ be bounded on $I:=[a,b]$ such that $f(x) \leq g(x)\ \forall\ x \in I$, and let $\mathcal{P}:=(x_0,x_1,\dots, x_n)$ be a partition of $[a,b]$ where
+				\[a=x_0<x_1<\dots<x_n=b\]
+				Let $M_{i_1}:=\sup \{f(x): x \in [x_{i-1},x_i]\}$, and let $m_{i_1}:=\inf \{f(x): x \in [x_{i-1},x_i]\}$, and $M_{i_2} := \sup \{g(x): x \in [x_{i-1},x_i]\}$ and $m_{i_2}:=\inf\{g(x):x\in[x_{i-1},x_i]\}$. Then since $f(x) \leq g(x)$, we know that $\sup f(x) \leq \sup g(x)$ and that $\inf f(x) \leq \inf g(x)$. This in turn means that
+				\[\sup \{f(x): x \in [x_{i-1},x_i]\} \leq \sup \{g(x): x \in [x_{i-1},x_i]\}\]
+				and
+				\[\inf \{f(x): x \in [x_{i-1},x_i]\} \leq \inf \{g(x): x \in [x_{i-1},x_i]\}\]
+				which implies that $M_{i_1} \leq M_{i_2}$ and $m_{i_1} \leq m_{i_2}$ for $i=1,2, \dots, n$.
+				\\\\Then we have the following:
+				\[L(f,\mathcal{P}) = \sum_{i=1}^{n} m_{i_1} (x_i-x_{i-1}) \leq \sum_{i=1}^{n} m_{i_2} (x_i-x_{i-1})= L(g,\mathcal{P})\]
+				So $L(f,\mathcal{P}) \leq L(g,\mathcal{P})$ and
+				\[U(f,\mathcal{P})=\sum_{i=1}^{n} M_{i_1} (x_i-x_{i-1}) \leq \sum_{i=1}^{n} M_{i_2}(x_i-x_{i-1})=U(g,\mathcal{P})\]
+				So $U(f,\mathcal{P}) \leq U(g,\mathcal{P})$.
+				\\\\Now, we have that
+				\[L(f)=\sup \{L(f,\mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \leq \sup \{L(g,\mathcal{P}): \mathcal{P} \in \mathscr{P}\}=L(g)\]
+				And so $L(f) \leq L(g)$. Also,
+				\[U(f)=\inf \{U(f,\mathcal{P}): \mathcal{P} \in \mathscr{P}\} \leq \inf \{U(g,\mathcal{P}): \mathcal{P} \in \mathscr{P}\}=U(g)\]
+				And thus $U(f) \leq U(g)$.\\\\
+				$\therefore$ If $f(x) \leq g(x)$, then $L(f) \leq L(g)$ and $U(f) \leq U(g)$.
+			\end{proof}
+			\item[5.] Let $f,g,h$ be bounded functions on $I:=[a,b]$ such that $f(x) \leq g(x) \leq h(x)$ for all $x \in I$. Show that if $f$ and $h$ are Darboux integrable and if $\displaystyle\int_{a}^{b} f=\displaystyle\int_{a}^{b} h$, then $g$ is also Darboux integrable with $\displaystyle\int_{a}^{b} g = \displaystyle\int_{a}^{b} f$.
+			
+			\begin{proof}
+				Let $f,g,h:[a,b] \to \R$ be bounded functions such that $f(x) \leq g(x) \leq h(x)\ \forall\ x \in [a,b]$, and suppose $f$ and $h$ are both Darboux integrable, and $\displaystyle\int_{a}^{b} f = \displaystyle\int_{a}^{b} h$. We want to show that $g$ is also Darboux integrable and that $\displaystyle\int_{a}^{b} g = \displaystyle\int_{a}^{b} f$.
+				\\\\Since $f$ and $h$ are Darboux integrable, we know that $U(f)=L(f)$ and $U(h)=L(h)$. Thus by the theorem posed in \textit{Problem 3}, we know that $U(f) \leq U(h)$ and $L(f) \leq L(h)$. We also know that $L(f)=U(f)=\displaystyle\int_{a}^{b} f$ and that $L(h)=U(h)=\displaystyle\int_{a}^{b} h=\int_{a}^{b} f$.\\\\
+				Again, by \textit{Problem 3}, we have that $L(f) \leq L(g) \leq L(h)$ and $U(f) \leq U(g) \leq U(h)$. Thus we have that $\displaystyle\int_{a}^{b} f \leq L(g) \leq \int_{a}^{b} f$ and $\displaystyle\int_{a}^{b} f \leq U(g) \leq \int_{a}^{b} f$, which thus yields that $L(g) = \int_{a}^{b} f$, and that $U(g) =\int_{a}^{b} f$. Hence $L(g)=U(g)=\displaystyle\int_{a}^{b} f$.
+				\\\\$\therefore$ $g$ is Darboux integrable and $\displaystyle\int_{a}^{b} g = \int_{a}^{b} f$.
+			\end{proof}
+			\item[6.] Let $f$ be defined on $[0,2]$ by $f(x):=1$ if $x \neq 1$ and $f(1) := 0$. Show that the Darboux integral exists and find its value.
+			\begin{proof}
+				Let $f(x):=\begin{cases}
+				1, &x \in [0,2]\setminus\{1\} \\
+				0, &x=1
+				\end{cases}$
+				\\Thus $f$ is bounded on $[0,2]$ Now, let $g:[0,2] \to \R$ be given by $g(x):=1$. Then since $g$ is a constant function, we know that $g$ is continuous on $[0,2]$, and thus by \textit{Theorem 7.2.7}, $g \in \mathcal{R}[0,2]$.
+				\\\\So, $\displaystyle\int_{0}^{2} 1\ dx= 2-0=2$. And since $f(x)=g(x)\ \forall\ x \in [0,2]\setminus\{1\}$, we have that $f \in \mathcal{R}[0,2]$. Also, $\displaystyle\int_{0}^{2} f=\int_{0}^{2} g=2$.
+				\\\\$\therefore$ By the \textit{Equivalence Theorem}, since $f\in\mathcal{R}[0,2]$, $f$ is Darboux integrable and $\displaystyle\int_{0}^{2} f =2$.\\
+			\end{proof}
+			\item[7.]
+			\begin{enumerate}
+				\item[a.] Prove that if $g(x):=0$ for $0 \leq x \leq \frac{1}{2}$ and $g(x):=1$ for $\frac{1}{2} < x \leq 1$, then the Darboux integral of $g$ on $[0,1]$ is equal to $\frac{1}{2}$.
+				\begin{proof}
+					Let $g(x):=\begin{cases}
+					0, &0 \leq x \leq \frac{1}{2} \\
+					1, &\frac{1}{2} < x \leq 1
+					\end{cases}$. We want to show that the Darboux integral of $g$ on $[0,1]$ is equal to $\frac{1}{2}$.
+					\\\\Since $g$ on the interval $\left[0,\frac{1}{2}\right]$ is a constant function, we know that $g$ is continuous on that interval, and is thus Riemann integrable, whose evaluation yields $\displaystyle\int_{0}^{\frac{1}{2}} g = \int_{0}^{\frac{1}{2}} 0 = 0$. 
+					\\\\Now, let $\varphi(x):= 1\ \forall\ x \in \left[\frac{1}{2},1\right]$. Then we have that $\varphi$ is a constant function and is thus continuous, and again by \textit{Theorem 7.2.7}, $\varphi$ is thus Riemann integrable, whose evaluation yields $\displaystyle\int_{\frac{1}{2}}^{1} 1 = 1-\frac{1}{2}=\frac{1}{2}$. Thus, $g=\varphi$ except at $\frac{1}{2}$. Hence $g$ is integrable on the interval $[0,1]$, and evaluates to $\displaystyle\int_{0}^{1} g = \int_{0}^{\frac{1}{2}} g + \int_{\frac{1}{2}}^{1} g = \frac{1}{2}$. Thus, by the \textit{Equivalence Theorem}, we have that $g$ is Darboux integrable.
+				\end{proof}
+				\item[b.] Does the conclusion hold if we change the value of $g$ at the point $\frac{1}{2}$ to $13$?
+				\\\\If we were to change $g$ at the one point then Riemann integrability is not affected, thus if $g(\frac{1}{2})=13$, then $g$ remains integrable on $[0,1]$ and $\displaystyle\int_{0}^{1} g = \frac{1}{2}$. Thus, $g$ is still Darboux integrable on $[0,1]$ with $\displaystyle\int_{0}^{1} g = \frac{1}{2}$.\\
+			\end{enumerate}
+			\item[9.] Let $f_1$ and $f_2$ be bounded functions on $[a,b]$. Show that $L(f_1)+L(f_2) \leq L(f_1 + f_2)$.
+			\begin{proof}
+				Consider the partitions $\mathcal{P}_1$ of $f_1$ on $[a,b]$, and  $\mathcal{P}_2$ of $f_2$ on $[a,b]$, and let $\mathcal{P}:=\mathcal{P}_1 \cup \mathcal{P}_2$ of $[a,b]$, where $\mathcal{P}:=(x_0,x_1,\dots,x_n)$ such that
+				\[a=x_0<x_1<\dots<x_n=b\]
+				
+				and let $m_{i_1} := \inf \{f_1(x): x \in [x_{i-1},x_i]\}$, $m_{i_2} := \inf \{f_2(x):x \in [x_{i-1},x_i]\}$, and let $m_{i_3}:=\inf \{f_1(x)+f_2(x):x \in [x_{i-1},x_i]\}$.
+				\\\\We note that $m_{i_1} + m_{i_2} \leq f_1(x)+f_2(x)\ \forall\ x \in [x_{i-1},x_i]$.
+				\\\\Then we have the following:
+				\begin{align*}
+					L(f_1,\mathcal{P}) + L(f_2,\mathcal{P}) &= \sum_{i=1}^{n} m_{i_1} (x_i-x_{i-1}) + \sum_{i=1}^{n} m_{i_2} (x_i-x_{i-1}) \\
+					&= \sum_{i=1}^{n} (m_{i_1}+m_{i_2}) (x_i-x_{i-1}) \\
+					&\leq \sum_{i=1}^{n} m_{i_3} (x_i-x_{i-1}), &\text{as was noted previously,} \\
+					&= L(f_1+f_2, \mathcal{P}) \leq L(f_1+f_2)
+				\end{align*}
+				Then we have that $\sup \{L(f_1,\mathcal{P}) + L(f_2, \mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \leq L(f_1+f_2)$. Thus,
+				\begin{align*}
+					L(f_1)+L(f_2) &=\sup\{L(f_1,\mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} + \sup \{L(f_2, \mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\ 
+					&\leq \sup \{L(f_1+f_2,\mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\
+					&=L(f_1+f_2)
+				\end{align*}
+				Thus we have that $L(f_1)+L(f_2) \leq L(f_1+f_2)$.
+			\end{proof}
+			\item[10.] Give an example to show that strict inequality can hold in the preceding exercise.
+			\\\\Consider the functions $f_1,f_2:[0,1] \to \R$ given by $f_1(x):=\begin{cases}
+			0, &x \in \Q \\
+			1, &x \in \R\setminus\Q
+			\end{cases}$, and $f_2(x):=\begin{cases}
+			1, &x \in \Q \\
+			0, &x \in \R\setminus\Q
+			\end{cases}$
+			\\Then we have that $L(f_1)=0$, and that $L(f_2)=0$, and thus $L(f_1)+L(f_2)=0$.
+			\\\\However, we now note that $(f_1+f_2)(x)=1\ \forall\ x \in [0,1]$, and thus we have that $L(f_1+f_2)=U(f_1+f_2)=\displaystyle\int_{0}^{1} 1 = 1 - 0=1$. Thus we have that $0=L(f_1)+L(f_2) < L(f_1+f_2) = 1$.
+		\end{enumerate}
+		\item Let $f(x)=x^2$ on $[1,3.5]$.
+		\begin{enumerate}
+			\item Find $L(f,P)$ and $U(f,P)$ when $P=\{1,2,3,3.5\}$.
+			\\\\Our terms are:
+			\[x_0:=1,\ \ x_1:=2,\ \ x_2:=3,\ \ x_3:=3.5\]
+			and our subintervals are
+			\[I_1:=[1,2],\ \ I_2:=[2,3],\ \ I_3:=[3,3.5]\]
+			So for $L(f,\mathcal{P})$ we have
+			\begin{align*}
+				L(f,\mathcal{P}) &= \sum_{i=1}^{n} m_i (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \inf \{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \inf \{x^2: x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \inf \{x^2: x \in [1,2]\}(2-1) \\
+				&+ \inf \{x^2: x \in [2,3]\}(3-2) \\
+				&+ \inf \{x^2: x \in [3,3.5]\}(3.5-3) \\
+				&= 1 \cdot 1 + 4 \cdot 1 + 9 \cdot \frac{1}{2} \\
+				&= 1 + 4 + \frac{9}{2} \\
+				&= \frac{19}{2}
+			\end{align*}
+			So $L(f,\mathcal{P}) = \frac{19}{2}$, and
+			\begin{align*}
+				U(f,\mathcal{P}) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup\{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup \{x^2: x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sup \{x^2: x \in [1,2]\}(2-1) \\
+				&+ \sup \{x^2: x \in [2,3]\}(3-2) \\
+				&+ \sup \{x^2: x \in [3,3.5]\}(3.5-3) \\
+				&= 4 \cdot 1 + 9 \cdot 1 + 12.25 \cdot \frac{1}{2} \\
+				&= 4 +9 + 6.125 \\
+				&= 19.125
+			\end{align*}
+			And so $U(f,\mathcal{P})=19.125$.\\\\
+			Thus $L(f,\mathcal{P})= \frac{19}{2}$ and $U(f,\mathcal{P})=19.125$.\\ 
+			\item Find $L(f,P)$ and $U(f,P)$ when $P=\{1,1.5,2,2.5,3,3.5\}$.
+			\\\\Our terms are:
+			\[x_0:=1,\ \ x_1:= 1.5,\ \ x_2:= 2,\ \ x_3:=2.5,\ \ x_4:=3,\ \ x_5:=3.5\]
+			and so our intervals are
+			\[I_1:=[1,1.5],\ \ I_2:=[1.5,2],\ \ I_3:=[2,2.5],\ \ I_4:=[2.5,3],\ \ I_5:=[3,3.5]\]
+			So we have for $L(f,\mathcal{P})$:
+			\begin{align*}
+				L(f,\mathcal{P}) &= \sum_{i=1}^{n} m_i (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \inf \{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \inf \{x^2: x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \inf \{x^2: x \in [1,1.5]\}(1.5-1) + \inf \{x^2: x \in [1.5,2]\}(2-1.5) \\
+				&+ \inf \{x^2: x \in [2,2.5]\}(2.5-2) + \inf \{x^2: x \in [2.5,3]\}(3-2.5) \\
+				&+ \inf \{x^2: x \in [3,3.5]\}(3.5-3) \\
+				&= 1 \cdot \frac{1}{2} + 2.25 \cdot \frac{1}{2} + 4 \cdot \frac{1}{2} + 6.25 \cdot \frac{1}{2} + 9 \cdot \frac{1}{2} \\
+				&= .5 +1.125 + 2+3.125 + 4.5 \\
+				&= 11.25
+			\end{align*}
+			thus $L(f,\mathcal{P})=11.25$ and 
+			\begin{align*}
+				U(f,\mathcal{P}) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup \{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup\{x^2: x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\
+				&= \sup \{x^2: x \in [1,1.5]\}(1.5-1) + \sup \{x^2: x \in [1.5,2]\}(2-1.5) \\
+				&+ \sup \{x^2: x \in [2,2.5]\}(2.5-2) + \sup \{x^2: x \in [2.5,3]\}(3-2.5) \\
+				&+ \sup \{x^2: x \in [3,3.5]\}(3.5-3) \\
+				&= 2.25 \cdot 0.5 + 4 \cdot 0.5 + 6.25 \cdot 0.5 + 9 \cdot 0.5 + 12.25 \cdot 0.5 \\
+				&= 16.875
+			\end{align*}
+			Thus $U(f,\mathcal{P}) = 16.875$.\\
+			Therefore $L(f,\mathcal{P}) = 11.25$ and $U(f,\mathcal{P})=16.875$.\\
+		\end{enumerate}
+		\item Use upper and lower Darboux sums to evaluate the following integrals.
+		\begin{enumerate}
+			\item $\displaystyle\int_{1}^{3} (2x+3)\ dx$
+			\\\\Let $\mathcal{P}_n :=\left(0, \frac{3}{n}, \frac{6}{n}, \dots,
+			 \frac{3n-1}{n}, 3\right)$. Then $\Delta x_i := \frac{3}{n}$. Since $2x+3$ is increasing on $[1,3]$,  we have that on $[x_{i-1},x_i] = \left[\frac{3i-1}{n}, \frac{3i}{n}\right]$, $M_i$ = occurs at the right endpoint = $f(\frac{3i}{n})=\frac{6i}{n} + 3$ and $m_i$ occurs at the left endpoint, $f\left(\frac{3i-1}{n}\right)=\frac{6i-2}{n}+3$.
+			 \\\\So, we also note that in order to get the correct answer, we must calculate $\displaystyle\int_{0}^{3} 2x+3\ dx - \int_{0}^{1} 2x+3\ dx$, and thus we choose $\mathcal{P}_1 := \left(0,\frac{1}{n},\dots,\frac{n-1}{n}, 1\right)$ with $\Delta x_{i_1} :=\frac{1}{n}$, with $M_{i_1}:= \frac{2i}{n}+3$, and $m_{i_1}:= \frac{2i-2}{n}+3$, and thus:
+			 \begin{align*}
+			 	U(f,\mathcal{P}_n)-U(f,\mathcal{P}_1) &= \sum_{i=1}^{n} M_i \Delta x_i - M_{i_1} \Delta x_{i_1} \\
+			 	&= \sum_{i=1}^{n} \left(\frac{6i}{n} + 3\right) \cdot \left(\frac{3}{n}\right) - \left[\left(\frac{2i}{n}+3\right)\cdot \left(\frac{1}{n}\right)\right]\\
+			 	&= \sum_{i=1}^{n} \frac{18i}{n^2} + \frac{9}{n} -
+			 	\frac{2i}{n^2}-\frac{3}{n}\\
+			 	&= \frac{18}{n^2}\ \sum_{i=1}^{n} i + \frac{9}{n} \sum_{i=1}^{n} 1 -\frac{2}{n^2}\ \sum_{i=1}^{n} i - \frac{3}{n} \sum_{i=1}^{n} 1\\
+			 	&= \frac{18}{n^2} \cdot \frac{n(n+1)}{2} + \frac{9\cancel{n}}{\cancel{n}} -\frac{2}{n^2}\cdot \frac{n(n+1)}{2}-\frac{3\cancel{n}}{\cancel{n}}\\
+			 	&= \frac{18n^2+18n}{2n^2} + 9 - \frac{2n^2+2n}{2n^2}-3 \\
+			 	&= \lim\limits_{n \to \infty} \frac{18n^2+18n}{2n^2} + 9 - \frac{2n^2+2n}{2n^2}-3 \\
+			 	&= 9+9-1-3 \\
+			 	&= 14 \\
+			 	&\geq U(f)
+			 \end{align*}
+			 and
+			 \begin{align*}
+			 	L(f,\mathcal{P}_n)-L(f,\mathcal{P}_1) &= \sum_{i=1}^{n} m_i \Delta x_i -m_{i_1} \Delta x_{i_1}\\
+			 	&= \sum_{i=1}^{n} \left(\frac{6i-2}{n}+3\right)\cdot \left(\frac{3}{n}\right) -\left[\left(\frac{2i-2}{n}+3\right)\cdot \left(\frac{1}{n}\right)\right]\\
+			 	&= \sum_{i=1}^{n} \frac{18i-6}{n^2} + \frac{9}{n} - \frac{2i-2}{n^2}-\frac{3}{n}\\
+			 	&= \sum_{i=1}^{n} \frac{6(3i-1)}{n^2} + \frac{9}{n}\ \sum_{i=1}^{n} 1 - \frac{2}{n^2}\ \sum_{i=1}^{n} (i-1) - \frac{3}{n}\ \sum_{i=1}^{n} 1 \\
+			 	&= \frac{6}{n^2}\ \left(3\ \sum_{i=1}^{n} i - \sum_{i=1}^{n} 1\right)+\frac{9\cancel{n}}{\cancel{n}} - \frac{2}{n^2} \cdot \frac{(n-1)((n-1)+1)}{2} -\frac{3\cancel{n}}{\cancel{n}}\\
+			 	&= \frac{6}{n^2}\cdot \left(\frac{3n(n+1)}{2}-n\right) + 9 - \frac{2}{n^2} \cdot \frac{n^2-n}{2} - 3\\
+			 	&= \frac{6}{n^2} \cdot \left(\frac{3n^2+3n}{2}-n\right) + 9 - \frac{2n^2-2n}{2n^2}-3\\
+			 	&= \frac{18n^2+18n}{2n^2} - \frac{6n}{n^2} +9 - \frac{2n^2-2n}{2n^2}-3 \\
+			 	&= \lim\limits_{n \to \infty} \frac{18n^2+18n}{2n^2} -\lim\limits_{n \to \infty} \frac{6}{n} + \lim\limits_{n \to \infty} 9 - \lim\limits_{n \to \infty} \frac{2n^2-2n}{2n^2} - \lim\limits_{n \to \infty} 3\\
+			 	&= 9-0+9-1-3 \\
+			 	&= 14 \\
+			 	&\leq L(f)
+ 			 \end{align*}
+ 			 So,
+ 			 \[14 \leq L(f) \leq U(f) \leq 14\]
+ 			 So $L(f)=U(f)=14$.\\
+			\item $\displaystyle\int_{0}^{2} (x^2+1)\ dx$
+			\\\\Let $\varepsilon > 0$ be given, and let $\mathcal{P}:= (0, \frac{2}{n}, \frac{4}{n}, \dots, \frac{2n-1}{n}, 2)$, and we note that $\Delta x_i := \frac{2}{n}$.
+			\\\\Since $f$ is increasing on $[0,2]$, then on $[x_{i-1},x_i] = \left[\frac{2i-1}{n}, \frac{2i}{n}\right]$, $M_i$ occurs at the right endpoint, and is thus $f\left(\frac{2i}{n}\right)=\frac{4i^2}{n^2}+1$. Also, $m_i$ occurs at the left endpoint and is thus $f\left(\frac{2i-1}{n}\right)=\left(\frac{4i^2-4i+1}{n^2}+1\right)$.
+			\begin{align*}
+				U(f,\mathcal{P}_n) &= \sum_{i=1}^{n} M_i \Delta x_i \\
+				&= \sum_{i=1}^{n} \left(\frac{4i^2}{n^2}+1\right) \cdot \left(\frac{2}{n}\right) \\
+				&= \sum_{i=1}^{n} \frac{8i^2}{n^3} + \frac{2}{n} \\
+				&= \frac{8}{n^3}\ \sum_{i=1}^{n} i^2 + \frac{2}{n}\ \sum_{i=1}^{n} 1 \\
+				&= \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{2 \cancel{n}}{\cancel{n}} \\
+				&= \frac{8}{n^3} \cdot \frac{2n^3+3n^2+n}{6} + 2 \\
+				&= \frac{16n^3+24n^2+8n}{6n^3} + 2 \\
+				&= \lim\limits_{n \to \infty}\frac{16n^3+24n^2+8n}{6n^3} + 2 \\
+				&= \frac{16}{6} + 2 \\
+				&= \frac{8}{3} + 2 \\
+				&= \frac{14}{3} \\
+				&\geq U(f)
+			\end{align*}
+			and
+			\begin{align*}
+				L(f,\mathcal{P}_n) &= \sum_{i=1}^{n} m_i \Delta x_i \\
+				&= \sum_{i=1}^{n} \left(\frac{4i^2-4i+1}{n^2}+1\right) \cdot \left(\frac{2}{n}\right) \\
+				&= \sum_{i=1}^{n} \frac{8i^2-8i+2}{n^3} + \frac{2}{n} \\
+				&= \sum_{i=1}^{n} \frac{2(2i-1)^2}{n^3} + \frac{2}{n}\ \sum_{i=1}^{n} 1 \\
+				&= \frac{8}{3}-\frac{2}{3n^2} + \frac{2 \cancel{n}}{\cancel{n}} \\
+				&= \lim\limits_{n \to \infty} \frac{8}{3} -\frac{2}{3n^2} + 2 \\
+				&= \frac{8}{3} -0 + 2 \\
+				&= \frac{8}{3} + 2 \\
+				&= \frac{14}{3} \\
+				&\leq L(f)
+			\end{align*}
+			So $\frac{14}{3} \leq L(f) \leq U(f) \leq \frac{14}{3}$. So $L(f)=U(f)=\frac{14}{3}$
+		\end{enumerate}
+		\item
+		\begin{enumerate}
+			\item Prove that if $f,g:[a,b] \to \R$ are bounded, then $U(f+g, \mathcal{P}) \leq U(f,\mathcal{P})+U(g,\mathcal{P})$ for every partition $\mathcal{P}$ of $[a,b]$.
+			\begin{proof}
+				Since $f$ and $g$ are bounded, we know that $\sup(f+g) \leq \sup (f) + \sup(g)$. Then $M_{f+g,i}:=\sup \{f+g:x \in [x_{i-1},x_i]\}$. And so $M_{f+g,i} \leq M_{f,i} + M_{g,i}$, and thus
+				\[U(f+g,\mathcal{P}):=\sum_{i=1}^{n} M_{f+g,i}\Delta x_i \leq \sum_{i=1}^{n} M_{f,i}\Delta x_i +\sum_{i=1}^{n} M_{g,i}\Delta x_i=U(f,\mathcal{P})+U(g,\mathcal{P})\]
+			\end{proof}
+			\item Find examples of bounded functions $f,g:[a,b] \to \R$ such that $U(f+g,\mathcal{P}) < U(f, \mathcal{P}) + U(g,\mathcal{P})$ for some partition of $[a,b]$.
+			\\\\Consider the functions $f,g:[a,b] \to \R$ given by $f(x):=\begin{cases}
+			1, &x \in \R\setminus\Q \\
+			0, &x \in \Q
+			\end{cases}$  and  $g(x):=\begin{cases}
+			-1, &x \in \R\setminus\Q \\
+			0, &x \in \Q
+			\end{cases}$
+			\\Thus, we have that $(f+g)(x) = 0\ \forall\  x$, and thus $U(f+g)=0$. However, we note that $U(f,\mathcal{P})+U(g,\mathcal{P})=1+0 = 1$. Thus, $0=U(f+g,\mathcal{P})<U(f,\mathcal{P})+U(g,\mathcal{P})=1$.
+		\end{enumerate}
+		\item Prove or justify, if true or provide a counterexample, if false.
+		\begin{enumerate}
+			\item Let $f$ be bounded on $[a,b]$. The upper and lower sums for $f$ form a bounded set.
+			\\\\This is a true statement.
+			\begin{proof}
+				Since $f$ is bounded, we know that $\exists\ m,M \st m \leq f(x) \leq M\ \forall\ x \in [a,b]$.\\
+				By the definitions of $L(f,\mathcal{P}),$ and $U(f,\mathcal{P})$, we have
+				\[L(f,\mathcal{P}):= \sum_{i=1}^n\inf (f)\cdot\Delta x_i\ \ \ \text{and}\ \ \ U(f,\mathcal{P}):= \sum_{i=1}^{n} \sup (f)\cdot\Delta x_i\]
+				This yields that 
+				\[m(b-a)\leq L(f,\mathcal{P})\leq U(f,\mathcal{P})\leq M(b-a)\]
+				So $L(f,\mathcal{P}),$ and $U(f,\mathcal{P})$ are bounded.
+			\end{proof}
+			\item Let $f$ be bounded on $[a,b]$. $f \in \mathcal{R}[a,b]$ if and only if its lower and upper sums are equal.
+			\\\\This is a false statement. Consider the function and partition given in \textit{Problem 1 (a)}:
+			Let $f(x):=|x|$ for $-1 \leq x \leq 2$. Calculate $L(f;P)$ and $U(f,P)$ for the following partition: $\mathcal{P}_1 :=(-1,0,1,2)$
+				\\\\Our terms are:
+				\[x_0:= -1,\ \ x_1:=0,\ \ x_2:=1,\ \ x_3:=2\]
+				and our intervals are:
+				\[I_1:=[-1,0],\ \ I_2:=[0,1],\ \ I_3:=[1,2]\]
+				thus $L(f,\mathcal{P}_1)$ is:
+				\begin{align*}
+				L(f,\mathcal{P}_1) &= \sum_{i=1}^{n} m_i(x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \left(\inf\{f(x):x \in [x_{i-1},x_i]\}\right) (x_i-x_{i-1}) \\
+				&= (\inf \{|x|:x \in [-1,0]\})(0-(-1))\\ 
+				&+ (\inf \{|x|: x \in [0,1]\})(1-0)\\
+				&+(\inf \{|x|: x\in [1,2]\})(2-1) \\
+				&= 0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1 \\
+				&= 0+0+1 \\
+				&= 1
+				\end{align*}
+				and
+				\begin{align*}
+				U(f,\mathcal{P}_1) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup \{f(x): x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\
+				&= \sum_{i=1}^{n} \sup \{|x|: x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\
+				&= (\sup \{|x|:x \in [-1,0]\})(0-(-1)) \\
+				&+ (\sup \{|x| : x \in [0,1]\})(1-0) \\
+				&+ (\sup \{|x|: x \in [1,2]\})(2-1) \\
+				&= 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 \\
+				&= 1 + 1 +2 \\
+				&= 4
+				\end{align*}
+				So, $L(f,\mathcal{P}_1)=1$ and $U(f,\mathcal{P}_1)=4$\\
+			\item Let $f$ be bounded on $[a,b]$. If $P$ and $Q$ are partitions of $[a,b]$, then $L(f,P) \leq U(f,Q)$.
+			\\\\This is a true statement by \textit{Lemma 7.4.3}:
+			\begin{lemma*}
+				Let $f:I\to\R$ be bounded. If $\mathcal{P}_1,\mathcal{P}_2$ are any two partitions of $I$, then $L(f;\mathcal{P}_1)\leq U(f;\mathcal{P}_2)$.
+			\end{lemma*}
+			\item When $\displaystyle\int_{a}^{b} f(x)\ dx$ exists, it is the unique number that lies between $L(f,P)$ and $U(f,P)$ for all partitions $P$ of $[a,b]$.
+			\\\\This is true since if $f \in \mathcal{R}[a,b]$, and if we let $I:=[a,b]$, then $f$ is Darboux integrable, by the \textit{Equivalence Theorem}. Then, by the definition of the Darboux integral, $U(f)=\inf \{U(f,\mathcal{P}) : \mathcal{P} \in \mathscr{P}(I)\}$ and $L(f) := \sup \{L(f,\mathcal{P}) : \mathcal{P} \in \mathscr{P}(I)\}$, and by \textit{Theorem 7.1.1}:
+			\begin{theorem*}
+				If $f \in \mathcal{R}[a,b]$, then the value of the integral is uniquely determined.
+			\end{theorem*}
+			, we have that the value $L$ of $\displaystyle\int_{a}^{b} f(x)\ dx=L$ is uniquely determined for all partitions $\mathcal{P} \in \mathscr{P}(I)$.
+			\\\\That is, this statement is a combination of \textit{Theorem 7.4.1} and \textit{Theorem 7.1.2}.\\
+			\item Let $f$ be bounded on $[a,b]$. Then $L(f,P) \leq \displaystyle\int_{a}^{b} f(x)\ dx \leq U(f,P)$.
+			\\\\This is false. Consider the Dirichlet function on the interval $[0,1]$: $f:[0,1] \to \R$ given by $f(x):=\begin{cases}
+			0, &x \in \Q \\
+			1, &x \in \R\setminus\Q
+			\end{cases}$\\
+			Then we have that $L(f,\mathcal{P}) = 0$ and $U(f,\mathcal{P}):=1$, so we have that $L(f,\mathcal{P}) \leq U(f,\mathcal{P})$, but we know that the Dirichlet function is not integrable. Thus we have that this is a false statement.\\
+			\item If $f \in \mathcal{R}[a,b]$, then for all $\varepsilon > 0$, there exists a partition $P$ of $[a,b]$ such that $L(f,P) > U(f,P)-\varepsilon$.
+			\\\\This is a true statement.
+			\begin{proof}
+				Let $f \in \mathcal{R}[a,b]$. Recall the \textit{Equivalence Theorem}:
+				\begin{theorem*}{\textbf{Equivalence Theorem}}
+					A function $f$ on $I=[a,b]$ is Darboux integrable if and only if it is Riemann integrable.
+				\end{theorem*}
+				Thus by the \textit{Equivalence Theorem}, $f$ is also Darboux integrable.
+				\\\\Also, recall the \textit{Integrability Criterion}:
+				\begin{theorem*}{\textbf{Integrability Criterion}}
+					Let $I:=[a,b]$ and let $f:I \to \R$ be a bounded function on $I$. Then $f$ is Darboux integrable on $I$ if and only if for each $\varepsilon > 0$ there is a partition $\mathcal{P}_\varepsilon$ of $I$ such that 
+					\[U(f;\mathcal{P}_\varepsilon)-L(f;\mathcal{P}_\varepsilon)<\varepsilon\]
+				\end{theorem*}
+				Note that we can rewrite the inequality as follows:
+				\begin{align*}
+					U(f,\mathcal{P}_\varepsilon) -L(f,\mathcal{P}_\varepsilon) < \varepsilon &\equiv -L(f,\mathcal{P}_\varepsilon) < \varepsilon-U(f,\mathcal{P}_\varepsilon) \\
+					&\equiv L(f,\mathcal{P}\varepsilon) > U(f,\mathcal{P}_\varepsilon) -\varepsilon
+				\end{align*}
+				Thus by the \textit{Integrability Criterion}, we have that $L(f,\mathcal{P})>U(f,\mathcal{P})-\varepsilon$.
+			\end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 3.tex

@@ -0,0 +1,521 @@
+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 3}
+\begin{document}
+	\maketitle
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 16 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+	\begin{enumerate}
+		\item 
+		\begin{enumerate}
+			\item[16.] If $f$ is continuous on $[a,b],a<b$, show that there exists $c \in [a,b]$ such that we have $\displaystyle\int_{a}^{b}f=f(c)(b-a)$. This result is sometimes called the \textit{Mean Value Theorem for Integrals}.
+			\begin{proof}
+				Let $m:=\inf \{f(x) : x \in [a,b]\}$ and $M:=\sup \{f(x): x \in [a,b]\}$. Then we know from \textit{Theorem 7.1.5 (c)} that
+				\[m(b-a) \leq \int_{a}^{b} f \leq M(b-a)\]
+				Then, dividing the inequality by $(b-a) > 0$, we have
+				\[m \leq \frac{\int_{a}^{b} f}{b-a} \leq M\]
+				By \textit{Bolzano's Theorem}, we can conclude that there exists $c \in [a,b] \st$
+				\[f(c):=\frac{\int_{a}^{b}f}{b-a}\]
+				which can be equivalently written as
+				\[\int_{a}^{b} f = f(c)(b-a)\]
+			\end{proof}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 19 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[19.] Suppose that $a>0$ and that $f \in \mathcal{R}[-a,a]$.
+			\begin{enumerate}
+				\item[(a)] If $f$ is \textit{even} (that is, if $f(-x)=f(x)$ for all $x \in [0,a]$), show that $\displaystyle\int_{-a}^{a}f=2\int_{0}^{a}f$.
+				\begin{proof}
+					Since $f$ is even, we have
+					\begin{align*}
+						\int_{-a}^{b} f &= \int_{-a}^{0} f(x)\ dx + \int_{0}^{a} f(x)\ dx \\
+						&= -\int_{a}^{0} f(-y)\ dy + \int_{0}^{a} f(x)\ dx
+					\end{align*}
+					where $y=-x$ for the first integral. Thus $x \mapsto -y,\ -a\mapsto a,\ 0 \mapsto 0$.
+					\begin{align*}
+						&= -\int_{a}^{0} f(y)\ dy + \int_{0}^{a} f(x)\ dx &\text{($f$ is even so $f(-y)=f(y)$} \\
+						&= \int_{0}^{a} f(y)\ dy + \int_{0}^{a} f(x)\ dx
+					\end{align*}
+					since flipping the limits of integration changes the sign of the integral
+					\begin{align*}
+						&= 2 \int_{0}^{a} f(x) \\
+						&= 2 \int_{0}^{a} f
+					\end{align*}
+				\end{proof}
+				\item[(b)] If $f$ is \textit{odd} (that is, if $f(-x)=-f(x)$ for all $x \in [0,a]$), show that $\int_{-a}^{a} f=0$.
+				\begin{proof}
+					Since $f$ is odd, we have
+					\begin{align*}
+						\int_{-a}^{a} f &= \int_{-a}^{0} f(x)\ dx + \int_{0}^{a} f(x)\ dx \\
+						&= -\int_{a}^{0} f(-y)\ dy + \int_{0}^{a} f(x)\ dx
+					\end{align*}
+					where $y=-x$, thus giving us $x \mapsto -y,\ -a \mapsto a,\ 0 \mapsto 0$.
+					\begin{align*}
+						&= -\int_{a}^{0} (-f(y))\ dy + \int_{0}^{a} f(x)\ dx &\text{since $f$ is odd, $f(-x)=-f(x)$} \\
+						&= \int_{0}^{a} (-f(y))\ dy + \int_{0}^{a} f(x)\ dx \\
+						&= -\int_{0}^{a} f(y)\ dy + \int_{0}^{a} f(x)\ dx
+					\end{align*}
+					since flipping the limits of integration changes the sign of the integral,
+					\begin{align*}
+						&= 0
+					\end{align*}
+					since both integrals cancel each other out. 
+				\end{proof}
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 20 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[20.] If $f$ is continuous on $[-a,a]$, show that $\int_{-a}^{a}f(x^2)dx=2\int_{0}^{a}f(x^2)dx$.
+			\begin{proof}
+				Since $f$ is continuous on $[-a,a]$, and $x^2$ is a continuous function, we know that $f(x^2)$ is a continuous function since the composition of continuous functions is continuous. That is, $f(x^2)$ is a continuous function since it is the composition of the continuous functions$f(x)$, and $x \mapsto x^2$. 
+				\\\\Let $g:[a,b] \to \R$ be given by $g(x):=f(x^2)$, then $g$ is also continuous as it is a composition of the continuous functions $x \mapsto f(x)$ and $x \mapsto x^2$. Notice, however, that $g(-x)=f((-x)^2)=f(x^2)=g(x)$. This means that $g$ is an even function.
+				\\\\So by the preceding problem, we know that
+				\[\int_{-a}^{a} g(x)\ dx = 2\int_{0}^{a} g(x)\ dx\]
+				Therefore
+				\[\int_{-a}^{a} f(x^2)\ dx = 2\int_{0}^{a} f(x^2)\ dx\]
+			\end{proof}
+		\end{enumerate}
+		\item 
+		\begin{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[3.] If $g(x):=x$ for $|x| \geq 1$ and $g(x):=-x$ for $|x|<1$ and if $G(x):=\frac{1}{2}|x^2-1|$, show that $\int_{-2}^{3}g(x)dx=G(3)-G(-2)=\frac{5}{2}$. Also sketch the graphs of $g$ and $G$.
+			\begin{proof}
+				Let $g(x):=\begin{cases}
+				x, &|x| \geq 1 \\
+				-x, &|x| < 1
+				\end{cases}$, and let $G(x):=\frac{1}{2} \abs{x^2-1}$. We want to show that $\int_{-2}^{3} g(x)\ dx=G(3)-G(-2)=\frac{5}{2}$.
+				\\\\Notice that since $G$ is a composition of continuous functions, namely $|x^2-1|$ and $x^2$, we know that $G$ is also continuous. Also, notice that $g$ has a finite number of discontinuities, namely at $x=-1$ and $x=1$. Thus, $g \in \mathcal{R}[-2,3]$.
+				\\\\Lastly, note that
+				\[G'(x):=\begin{cases}
+				x, &|x| \geq 1 \\
+				-x, &|x| < 1
+				\end{cases}=g(x),\ \forall\ x \in [-2,3]\setminus\{-1,1\}\]
+				Thus, by the Fundamental Theorem of Calculus, we have
+				\[\int_{-2}^{3} g(x)\ dx = G(3)-G(-2) = \frac{1}{2} |9-1| - \frac{1}{2}|4-1| = \frac{8}{2}-\frac{3}{2}=\frac{5}{2}\]
+			\end{proof}
+			\begin{tikzpicture}
+			\begin{axis}[
+				axis x line=middle, axis y line=middle,
+				ymin=-2, ymax=3, ylabel={$g(x)$},
+				xmin=-2, xmax=3, xlabel={$x$},
+				domain=-2:3
+			]
+			
+			\addplot[blue,thick][domain=-2:-1]{x};
+			\addplot[blue,thick][domain=-1:1]{-x};
+			\addplot[blue,thick][domain=1:3]{x};
+			\end{axis}
+			\end{tikzpicture}
+			\begin{tikzpicture}
+				\begin{axis}[
+				axis x line=middle, axis y line=middle,
+				ymin=-2, ymax=3, ylabel={$G(x)$},
+				xmin=-2, xmax=3, xlabel={$x$},
+				domain=-2:3
+				]
+				
+				\addplot[blue,thick][domain=-2:3, samples=150]{1/2 * abs(x^2-1)};
+				\end{axis}
+			\end{tikzpicture}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 9 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
+			\item[9.] Let $f \in \mathcal{R}[a,b]$ and define $F(x):=\displaystyle\int_{a}^{x}f$ for $x \in [a,b]$.
+			\begin{enumerate}
+				\item[(a)] Evaluate $G(x):=\displaystyle\int_{c}^{x}f$ in terms of $F$, where $c \in [a,b]$.
+				\begin{align*}
+					G(x)&=\int_{c}^{x} f \\
+					&=\int_{a}^{c} f + \int_{c}^{x} f - \int_{a}^{c} f \\
+					&= \int_{a}^{x} f-\int_{a}^{c} f \\
+					&= F(x)-F(c)
+				\end{align*}
+				\item[(b)] Evaluate $H(x):=\displaystyle\int_{x}^{b} f$ in terms of $F$.
+				\begin{align*}
+					H(x)&=\int_{x}^{b} f \\
+					&= \int_{a}^{x} f + \int_{x}^{b} f- \int_{a}^{x} f \\
+					&= F(b)-F(x)
+				\end{align*}
+				\item[(c)] Evaluate $S(x):=\displaystyle\int_{x}^{\sin x} f$ in terms of $F$.
+				\begin{align*}
+					S(x)&= \int_{x}^{\sin x} f \\
+					&= \int_{a}^{x} f + \int_{x}^{\sin x} f - \int_{a}^{x} f \\
+					&= \int_{a}^{\sin x} f - \int_{a}^{x} f \\
+					&= F(\sin x) - F(x)
+				\end{align*}
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 11 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[11.] Find $F'(x)$ when $F$ is defined on $[0,1]$ by:
+			\begin{enumerate}
+				\item[(a)] $F(x):=\displaystyle\int_{0}^{x^2} (1+t^3)^{-1}\ dt$.
+				\\\\Since $x^2$ is continuous and differentiable on $[0,1]$, we can use \textit{Leibniz's Rule} to find $F'(x)$, where $f(x)=\frac{1}{1+x^3}$, $h(x):=x^2$, and $g(x):=0$. So,
+				\begin{align*}
+					F'(x)&= f(h(x))\cdot h'(x)-f(g(x))\cdot g'(x) \\
+					&= \frac{1}{1+(x^2)^3}\cdot 2x - \frac{1}{1+(0)^3}\cdot 0 \\
+					&= \frac{2x}{1+x^6}-0 \\
+					&= \frac{2x}{1+x^6}
+				\end{align*}
+				\item[(b)] $F(x):=\displaystyle\int_{x^2}^{x} \sqrt{1+t^2}\ dt$.
+				\\\\Since both $x$ and $x^2$ are continuous and differentiable on $[0,1]$, we can use \textit{Leibniz's Rule} to find $F'(x)$, where $f(x):=\sqrt{1+x^2},\ h(x):=x,$ and $g(x):=x^2$. So, we must first rewrite $F(x)$ as
+				\[F(x):=\int_{x^2}^{x} \sqrt{1+x^2}\ dx = \int_{0}^{x} \sqrt{1+x^2}\ dx - \int_{0}^{x^2} \sqrt{1+x^2}\ dx\]
+				\begin{align*}
+					F'(x)&=f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x) \\
+					&= \sqrt{1+x^2} \cdot 1 - \sqrt{1+(x^2)^2} \cdot 2x \\
+					&= \sqrt{1+x^2} - 2x \cdot \sqrt{1+x^4}
+				\end{align*}
+			\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[12.] Let $f:[0,3] \to \R$ be defined by $f(x):=x$ for $0 \leq x < 1$, $f(x):=1$ for $1 \leq x < 2$ and $f(x):=x$ for $2 \leq x \leq 3$. Obtain formulas for $F(x):=\int_{0}^{x} f$ and sketch the graphs of $f$ and $F$. Where is $F$ differentiable? Evaluate $F'(x)$ at all such points.
+			\\\\Let $f(x):=\begin{cases}
+			x, &0 \leq x < 1 \\
+			1, &1 \leq x < 2 \\
+			x, &2 \leq x \leq 3
+			\end{cases}$
+			\\Then we have the following: When $x \in [0,1)$:
+			\begin{align*}
+				F(x) &= \int_{0}^{x} f(t)\ dt \\
+				&= \int_{0}^{x} t\ dt \\
+				&= \frac{x^2}{2}
+			\end{align*}
+			When $x \in [1,2)$:
+			\begin{align*}
+				F(x) &= \int_{0}^{x} f(t)\ dt \\
+				&= \int_{0}^{1} t\ dt + \int_{1}^{x} 1\ dt \\
+				&= \frac{1}{2} + (x-1) \\
+				&= x-\frac{1}{2}
+			\end{align*}
+			When $x \in [2,3]$:
+			\begin{align*}
+				F(x) &= \int_{0}^{x} t\ dt \\
+				&= \int_{0}^{1} t\ dt + \int_{1}^{2} 1\ dt + \int_{2}^{3} t\ dt \\
+				&= \frac{1}{2} + 1 + \left(\frac{x^2}{2}-\frac{2^2}{2}\right) \\
+				&= \frac{x^2}{2} - \frac{1}{2}
+			\end{align*}
+			Therefore, we have
+			\[F(x)=\begin{cases}
+			\frac{x^2}{2}, &0 \leq x < 1 \\
+			x-\frac{1}{2}, &1 \leq x < 2 \\
+			\frac{x^2}{2} - \frac{1}{2}, &2 \leq x \leq 3 
+			\end{cases}\]
+			\begin{tikzpicture}
+				\begin{axis}[
+					axis x line*=bottom, axis y line*=left,
+					ymin=0, ymax=4, ytick distance=1, ylabel={$F(x)$},
+					xmin=0, xmax=3, xtick distance=1
+				]
+				
+				\addplot[blue,thick][domain=0:1]{x^2/2};
+				\addplot[blue,thick][domain=1:2]{x-1/2};
+				\addplot[blue,thick][domain=2:3]{x^2/2 - 1/2};
+				\end{axis}
+			\end{tikzpicture}\\
+			\begin{tikzpicture}
+				\begin{axis}[
+				axis x line*=bottom, axis y line*=left,
+				ymin=0, ymax=4, ytick distance=1, ylabel={$f(x)$},
+				xmin=0, xmax=3, xtick distance=1
+				]
+				
+				\addplot[blue,thick][domain=0:1]{x};
+				\addplot[blue,thick][domain=1:2]{1};
+				\addplot[blue,thick][domain=2:3]{x};
+				\end{axis}
+			\end{tikzpicture}
+			\\$F$ is definitely differentiable at points $x \in (0,1) \cup (1,2) \cup (2,3)$ since at those points, $f$ is equal to a polynomial. So, we must now check for the differentiability of $f$ at $x=1$ and $x=2$.
+			\[\limx{x}{1^-}\frac{F(x)-F(1)}{x-1}=\limx{x}{1^-}\frac{\frac{x^2}{2}-\frac{1}{2}}{x-1}=\frac{1}{2} \limx{x}{1^-}\frac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)}}=\frac{1}{2} \cdot (1+1) = 1\]
+			\[\limx{x}{1^+}\frac{F(x)-F(1)}{x-1}=\limx{x}{1^+}\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{x-1} = 1\]
+			Therefore, $F$ is differentiable at $x=1$ and $F'(x)=1$. As for when $x=2$,
+			\[\limx{x}{2^-}\frac{F(x)-F(2)}{x-1}=\limx{x}{2^-}\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{x-2}=1\]
+			\[\limx{x}{2^+}\frac{F(x)-F(2)}{x-2}=\limx{x}{2^+}\frac{\left(\frac{x^2}{2}-\frac{1}{2}\right)-\frac{1}{2}}{x-2}=2\]
+			Therefore, $F$ is not differentiable at $x=2$. Thus,
+			\[F'(x):=\begin{cases}
+			x, &0 \leq x < 1 \\
+			1, &1 \leq x < 2 \\
+			x, &2 < x \leq 3
+			\end{cases}\]
+			And notice that $F'(x)=f(x)$ for $x \in [0,1] \setminus \{2\}$
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 13 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[13.] The function $g$ is defined on $[0,3]$ by $g(x):=-1$ if $0 \leq x < 2$ and $g(x):=1$ if $2 \leq x \leq 3$. Find the indefinite integral $G(x)=\int_{0}^{x} g$ for $0 \leq x \leq 3$, and sketch the graphs of $g$ and $G$. Does $G'(x)=g(x)$ for all $x \in [0,3]$?
+			\[g(x):=\begin{cases}
+			-1, &0 \leq x < 2 \\
+			1, & 2 \leq x \leq 3
+			\end{cases}\]
+			Then we have the following: when $x \in [0,2)$:
+			\begin{align*}
+				G(x) &= \int_{0}^{x} g(t)\ dt \\
+				&= \int_{0}^{x} -1\ dt \\
+				&= -x
+			\end{align*}
+			When $x \in [2,3]$:
+			\begin{align*}
+				G(x)&= \int_{0}^{x} g(t)\ dt \\
+				&= \int_{0}^{2} -1\ dt + \int_{2}^{x} 1\ dt \\
+				&= -2+x-2 \\
+				&= x-4
+			\end{align*}
+			Therefore,
+			\[F(x)=\begin{cases}
+			-x, &0 \leq x < 2 \\
+			x-4, &2 \leq x \leq 3
+			\end{cases}\]
+			\\\\\begin{tikzpicture}
+				\begin{axis}[
+					axis x line=middle, axis y line=left,
+					ymin=-2, ymax=1, ytick distance=1, ylabel={$g(x)$},
+					xmin=0, xmax=3, xtick distance=1
+				]
+				
+				\addplot[red,thick][domain=0:2]{-1};
+				\addplot[red,thick][domain=2:3]{1};
+				\end{axis}
+			\end{tikzpicture}\\
+			\begin{tikzpicture}
+				\begin{axis}[
+					axis x line=middle, axis y line=middle,
+					ymin=-3, ymax=3, ytick distance=1, ylabel={$G(x)$},
+					xmin=0, xmax=3, xtick distance=1
+				]
+				
+				\addplot[blue,thick][domain=0:2]{-x};
+				\addplot[blue,thick][domain=2:3]{x-4};
+				\end{axis}
+			\end{tikzpicture}
+			\\\\Now, if it is possible, let $G'(x)=g(x)\ \forall\ x \in [0,3]$. Then
+			\[\limx{x}{2^-}\frac{G(x)-G(2)}{x-2}=\limx{x}{2^-}\frac{-x+2}{x-2} = -1\]
+			\[\limx{x}{2^+}\frac{G(x)-G(2)}{x-2}=\limx{x}{2^+}\frac{x-2}{x-2}=1\]
+			So,
+			\[\limx{x}{2^-} \frac{G(x)-G(2)}{x-2}\neq \limx{x}{2^+}\frac{G(x)-G(2)}{x-2}\]
+			Thus we have that the limit does not exist, and hence $G$ is not differentiable at $x=2$. Thus $G'(x) \neq g(x)$ for some $x \in [0,3]$.\\
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 16 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[16.] If $f:[0,1] \to \R$ is continuous and $\int_{0}^{x} f=\int_{x}^{1} f$ for all $x \in [0,1]$, show that $f(x)=0$ for all $x \in [0,1]$.
+			\begin{proof}
+				Let $F(x)=\displaystyle\int_{0}^{x} f(t)\ dt$ for $x \in [0,1]$. $F$ is well defined since $f$ is continuous on $[0,1]$, and thus is also integrable on $[0,1]$. $F$ is also differentiable since $f$ is continuous and $F'(x)=f(x)\ \forall\ x \in [0,1]$. So,
+				\begin{align*}
+					\int_{0}^{x} f = \int_{x}^{1} f &\Leftrightarrow \int_{0}^{x} f=\int_{0}^{1} f -\int_{0}^{x} f \\
+					&\Leftrightarrow 2 \int_{0}^{x} f = \int_{0}^{1} f \\
+					&\Leftrightarrow 2F(x)=F(1)
+				\end{align*}
+				And differentiating the last relation with respect to $x$, we get
+				\[2F'(x)=0 \Leftrightarrow F'(x)=0 \Leftrightarrow f(x)=0,\ \forall\ x \in [0,1]\]
+			\end{proof}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 18c %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+			\item[18.] Use the Substitution Theorem 7.3.8 to evaluate the following integral:
+			\begin{enumerate}
+				\item[(c)] $\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}} dt$
+				\\\\Let $\phi(t)=1+\sqrt{t}$ for $t \in [1,4]$, and let $f(u)=\sqrt{u}$ for $u \in [2,3]$. $f$ is continuous on $[2,3]$ and $\phi$ has a continuous derivative (namely, $\phi'(t)=\frac{1}{2\sqrt{t}}$) on $[1,4]$. Thus we have
+				\begin{align*}
+					\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}}\ dt &= 2\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}}\ dt \\
+					&= 2 \int_{1}^{4} \phi'(t)\cdot f(\phi(t))\ dt \\
+					&= 2 \int_{\phi(1)}^{\phi(4)} f(u)\ dt &\text{by the \textit{Substitution Theorem}} \\
+					&= 2 \int_{\phi(1)}^{\phi(4)} \sqrt{u}\ du \\
+					&= 2 \int_{2}^{3} \sqrt{u}\ du \\
+					&= 2 \cdot \left.\frac{2}{3}u^{\frac{3}{2}}\right|_{u=2}^{u=3} \\
+					&= \frac{4}{3}\left(3^{\frac{3}{2}}-2^{\frac{3}{2}}\right)
+				\end{align*}
+			\end{enumerate}
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
+		\item Let $f:[0,1] \to \R$ given by $f(x)=\begin{cases}
+		\frac{1}{n}\ &\text{if } x \in \left(\frac{1}{n+1},\frac{1}{n}\right] \\
+		0\ &\text{if }x=0
+		\end{cases}$. Sketch the graph of $f$ and show that $f \in \mathcal{R}[0,1]$.\\
+		\begin{tikzpicture}
+			\begin{axis}[
+				axis x line=bottom, axis y line=left,
+				ymin=0, ymax=4, ytick distance= 1, ytick={1,2,3,4}, yticklabels={$\dots$,$\frac{1}{3}$, $\frac{1}{2}$, $1$},
+				xmin=0, xmax=4, xtick distance=1, xtick={1,2,3,4},
+				xticklabels={$\dots$,$\frac{1}{3}$, $\frac{1}{2}$, $1$}
+			]
+			
+			\addplot[blue,thick][domain=3:4]{4};
+			\addplot[blue,thick][domain=2:3]{3};
+			\addplot[blue,thick][domain=1:2]{2};
+			\addplot[blue,thick][mark=*] coordinates{(4,4)};
+			\addplot[blue,thick][mark=*,fill=white] coordinates{(3,4)};
+			\addplot[blue,thick][mark=*] coordinates{(3,3)};
+			\addplot[blue,thick][mark=*,fill=white] coordinates{(2,3)};
+			\addplot[blue,thick][mark=*] coordinates{(2,2)};
+			\addplot[blue,thick][mark=*,fill=white] coordinates{(1,2)};
+			\addplot[blue,thick][mark=*] coordinates{(0,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\\Since $f$ is monotone, by \textit{Theorem 7.2.8}, $f \in \mathcal{R}[0,1]$.
+		
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item
+		\begin{enumerate}
+			\item Give an example of two functions $f,g:[a,b] \to \R$ that are not Riemann integrable, but $fg \in \mathcal{R}[a,b]$.
+			\\\\Consider 
+			\[f(x):=\begin{cases}
+			1, &x \in \Q \\
+			0, &x \in \R\setminus\Q
+			\end{cases}\ \ \ \ \text{  and  }\ \ \ \ g(x):=\begin{cases}
+			0, &x \in \Q \\
+			1, &x \in \R\setminus\Q
+			\end{cases}\]
+			Since these are both Dirichlet and modified Dirichlet functions, we know that they are not Riemann integrable, however,
+			\[fg:=\begin{cases}
+			0, x \in \Q \\
+			0, x \in \R\setminus\Q
+			\end{cases}=0\]
+			and thus $fg$ is a constant function, which is Riemann integrable. Thus we have that $f,g \notin \mathcal{R}[a,b]$, but $fg \in \mathcal{R}[a,b]$.\\
+			\item Give an example of two functions $f,g:[a,b] \to \R$ where $f \in \mathcal{R}[a,b]$ and $g \notin \mathcal{R}[a,b]$, but $fg \in \mathcal{R}[a,b]$.\\\\
+			Consider
+			\[f(x):=0,\ \forall\ x \in [a,b],\ \ \ \ \text{  and  }\ \ \ \ g(x):=\begin{cases}
+			1, &x \in \Q \\
+			0, &x \in \R\setminus\Q
+			\end{cases}\]
+			Then since $f$ is a constant function, $f \in \mathcal{R}[a,b]$, and since $g$ is the Dirichlet function, we know that $g \notin \mathcal{R}[a,b]$. However,
+			\[fg=\begin{cases}
+			0, &x \in \Q \\
+			0, &x \in \R\setminus\Q
+			\end{cases}=0\]
+			And thus $fg$ is a constant function and is thus Riemann integrable. Thus we have that $f \in \mathcal{R}[a,b]$, $g \notin \mathcal{R}[a,b]$, and $fg \in \mathcal{R}[a,b]$.\\
+			\item Let $f:[a,b] \to \R$, $f \in \mathcal{R}[a,b]$. Let $F:[a,b] \to \R$ be given by $F(x)=\displaystyle\int_{a}^{x} f(t)dt$. Prove that $F$ is Lipschitz.
+			\begin{proof}
+				Since $f \in \mathcal{R}[a,b]$, $f$ is bounded; that is, there is some $M \st |f(x)| \leq M\ \forall\ x \in [a,b]$. Now, if $y < x$, we have
+				\[|F(x)-F(y)|=\abs{\int_{y}^{x} f(t)\ dt} \leq \int_{y}^{x} |f(t)|\ dt \leq \int_{y}^{x} M\ dt = M(x-y)=M|x-y|\]
+				Similarly, $|F(x)-F(y)| \leq M(y-x)=M|x-y|$ if $y > x$. So, we see that for any $x,y \in [a,b]$, if we let $K=M$, we have
+				\[|F(x)-F(y)| \leq K|x-y|\]
+%				Thus, $F$ is Lipschitz since the above statement is the definition of a Lipschitz function.
+			\end{proof}
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Let $f(t):=\begin{cases}
+		t\ &\text{for } 0 \leq t \leq 2 \\
+		3\ &\text{for } 2 < t \leq 4
+		\end{cases}$
+		\begin{enumerate}
+			\item Find an explicit expression for $F(x)=\int_{0}^{x}f(t)dt$.
+			\\\\When $x \in [0,2]$:
+			\begin{align*}
+				F(t)&= \int_{0}^{x} f(t)\ dt \\
+				&= \int_{0}^{x} t\ dt \\
+				&= \frac{x^2}{2}
+			\end{align*}
+			and when $x \in (2,4]$:
+			\begin{align*}
+				F(t)&= \int_{0}^{x} f(t)\ dt \\
+				&= \int_{0}^{2} t\ dt + \int_{2}^{x} 3\ dt \\
+				&= 2+3x-6 \\
+				&= 3x-4
+			\end{align*}
+			Thus,
+			\[F(x):=\begin{cases}
+			\frac{x^2}{2}, &0 \leq x \leq 2 \\
+			3x-4, &2 < t \leq 4
+			\end{cases}\]
+			\item Sketch $F$ and determine where $F$ is differentiable.\\\\
+			\begin{tikzpicture}
+				\begin{axis}[
+					axis x line=bottom, axis y line=left,
+					ymin=0, ymax=8, ytick distance=1,
+					xmin=0, xmax=4, xtick distance=1
+				]
+				
+				\addplot[blue,thick][domain=0:2]{x^2/2};
+				\addplot[blue,thick][domain=2:4]{3*x -4};
+				\end{axis}
+			\end{tikzpicture}
+			\\\\Based on the graph, we can tell that the only place in which $F$ is not differentiable is at $x=2$, which we can see as follows:
+			\[\limx{x}{2^-} \frac{F(x)-F(x)}{x-2} = \limx{x}{2^-} \frac{\frac{x^2}{2}-2}{x-2}=2\]
+			\[\limx{x}{2^+}\frac{F(x)-F(2)}{x-2}=\limx{x}{2^+} \frac{3x-4-2}{x-2}=\limx{x}{2^+} \frac{3x-2}{x-2} = 3\]
+			Since $\limx{x}{2^-} \frac{F(x)-F(2)}{x-2} \neq \limx{x}{2^+} \frac{F(x)-F(2)}{x-2}$, we have that $F$ is not differentiable when $x=2$.\\
+			\item Find formula for $F'(x)$ wherever $F$ is differentiable.
+			\\\\Since the only place in which $F$ is not differentiable is when $x=2$, we need only change one of the inequalities of $f$. So,
+			\[F'(x):=\begin{cases}
+			x, &0 \leq x < 2 \\
+			3, &2<x\leq 4
+			\end{cases}\]
+		\end{enumerate}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+		\item Prove or justify, if true or provide a counterexample, if false.
+		\begin{enumerate}
+			\item If $f \in \mathcal{R}[a,b]$, then $|f| \in \mathcal{R}[a,b]$ and $\displaystyle\int_{a}^{b} |f| \leq \abs{\int_{a}^{b}f}$.
+			\\\\This is false since the inequality is flipped from how it appears in \textit{Corollary 7.3.15} to the \textit{Composition Theorem}. Consider $f:[0,2\pi] \to R$ given by $f(x):=\sin x$. Then we have
+			\[\abs{\int_{0}^{2\pi} \sin x\ dx} = \abs{0} = 0 \leq 4=\int_{0}^{2\pi} \abs{\sin x}\ dx\]
+			and thus
+			\[\abs{\int_{0}^{2\pi} \sin x\ dx} \leq \int_{0}^{2\pi} \abs{\sin x}\ dx\]
+			\item If $f,g \in \mathcal{R}[a,b]$ and $f$ is continuous, then there exists a $c \in [a,b]$ such that $\displaystyle\int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx$.
+			\\\\This is a false statement. Consider $f,g:[-1,1] \to \R$ given by $f(x):=x+2$ and $g(x):=x$. Then we have the following:
+			\[\int_{-1}^{1} fg = \int_{-1}^{1} x^2+2x\ dx = \frac{2}{3} \neq 0\cdot (x+2)=f(c) \int_{-1}^{1} x\ dx\]
+			Thus $\nexists\ c \in [a,b] \st \displaystyle\int_{-1}^{1} f(x)g(x)=f(c) \cdot \int_{-1}^{1} g(x)\ dx$ for $f,g$ as given.
+			\item If $f \in C^1(\R)$, then $\frac{d}{dx}\displaystyle\int_{0}^{x}f(t)dt = \int_{0}^{x}\left[\frac{d}{dx}f(t)\right]dt$.\\\\
+			This is a false statement, consider the function $f(x):=\cos x$. Then,
+			\[\frac{d}{dx}\int_{0}^{x} \cos x\ dx = \frac{d}{dx} (\sin x)=\cos x\]
+			and
+			\[\int_{0}^{x}\left[\frac{d}{dx}\cos x\right]\ dx = \int_{0}^{x} -\sin x\ dx = \cos x-1\]
+			Hence we have that equality does not hold.\\
+			\item If $f'(x)=\sin x - \cos x$, then $f(x)=\displaystyle\int_{0}^{x} (\sin t - \cos t) dt$.
+			\\\\This is a false statement. Note the following:
+			\[\int_{0}^{x} (\sin t - \cos t)\ dt = (-\sin x - \cos x + 1)\]
+			but
+			\[f(x)=\int (\sin x - \cos x)\ dx = -\cos x- \sin x\]
+			Hence
+			\[-\sin x - \cos x + 1 \neq -\sin x - \cos x\]
+			\item If $f,g:[a,b] \to \R$ are such that $f,fg \in \mathcal{R}[a,b]$ and $f$ is strictly monotone on $[a,b]$, then $g \in \mathcal{R}[a,b]$?
+			\\\\This is a false statement. Consider $f,g:[-1,1] \to \R$ given by $f(x)=x$ and $g(x)=\frac{1}{x}$. Then both $f, fg \in \mathcal{R}[-1,1]$ since $x$ is integrable and $fg=x \cdot \frac{1}{x} = \frac{x}{x}=1$ is also integrable. However, since $\limx{x}{0^-} g(x)=-\infty$ and $\limx{x}{0^+} g(x)=\infty$, we have that $g(x) \notin \mathcal{R}[-1,1]$.
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 4.tex

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 4}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item Evaluate
+		\begin{enumerate}
+			\item $\displaystyle\lim\limits_{h \to 0} \frac{1}{h} \int_{x}^{x+h} \sqrt{t+1}\cos t\ dt$ \\
+			\begin{align*}
+				\limx{h}{0}\frac{1}{h} \int_{x}^{x+h} \sqrt{t+1}\cos t\ dt &= \limx{h}{0} \frac{\displaystyle\int_{0}^{x+h}\sqrt{t+1} \cos t\ dt - \int_{0}^{x} \sqrt{t+1}\cos t\ dt}{h} \\
+				&=\limx{h}{0} \frac{F(x+h)-F(x)}{h},\  \text{where $F(x):=\int_{0}^{x}\sqrt{t+1}\cos t\ dt$} \\
+				F'(x) &= \frac{d}{dx} \int_{0}^{x} \sqrt{t+1}\cos t\ dt \\
+				&\text{let } h(x)=x,\ g(x)=0,\text{ and } f(x):=\sqrt{x+1}\cos x\text{ then} \\
+				F'(x) &= f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x) \\
+				&= \sqrt{x+1}\cos x \cdot 1 -\sqrt{0+1}\cos 0 \cdot 0 \\
+				&= \sqrt{x+1}\cos x		
+			\end{align*}
+			Thus $F'(x)=\sqrt{x+1}\cos x$.\\
+			\item $\displaystyle\lim\limits_{x \to a} \frac{x}{x-a} \int_{a}^{x} t^2\ dt$ \\
+			\begin{align*}
+				\limx{x}{a} \frac{x}{x-a} \int_{a}^{x} t^2\ dt &= \limx{x}{a} \frac{x \cdot \displaystyle\int_{a}^{x} t^2\ dt}{x-a} \\
+				&= \lim\limits_{x \to a} \frac{x(x^2\cdot 1 -a^2\cdot 0)}{1-0} \\
+				&= a^3
+			\end{align*}
+			\item $\displaystyle\lim\limits_{x \to 0} \frac{1}{x} \int_{0}^{x} \sqrt{9 + t^2}\ dt$
+			\begin{align*}
+				\limx{x}{0} \frac{1}{x} \int_{0}^{x} \sqrt{9+t^2}\ dt &= \limx{x}{0} \frac{\displaystyle\int_{0}^{x} \sqrt{9+t^2}\ dt}{x} \\
+				&= \limx{x}{0} \frac{\sqrt{9+x^2}\cdot 1 - \sqrt{9+0^2}\cdot 0}{1}, &\text{by L'Hospital's Rule and Leibniz's Rule} \\
+				&= \limx{x}{0} \sqrt{9+x^2} \\
+				&= \sqrt{9} \\
+				&= 3
+			\end{align*}
+		\end{enumerate}
+		\item 
+		\begin{enumerate}
+			\item Show that $(x^2\sin x) /2$ is not an antiderivative of $x \cos x$.
+			\begin{proof}
+				We want to show that $\frac{x^2\sin x}{2} \neq \int x\cos x $ So, we note that we can use \textit{Theorem 7.3.17 Integration by Parts}. So, let $u=x$ and $dv=\cos x\ dx$. Then
+				\[du = u'dx=1\cdot dx = dx\]
+				and
+				\[v=\int \cos x\ dx = \sin x + C\]
+				for some arbitrary constant $C$. Then,
+				\begin{align*}
+					\int x \cos x\ dx &= uv - \int v\ du \\
+					&= x\sin x - \int \sin x\ dx \\
+					&= x\sin x - (-\cos x + C) \\
+					&= x\sin x + \cos x + C
+				\end{align*}
+				for some arbitrary constant $C$. And thus we have that the antiderivative of $x \cos x=x\sin x +\cos x$, which we note is \textit{not} equal to $\frac{x^2\sin x}{2}$. That is,
+				\[\int x \cos x = x \sin x +\cos x \neq \frac{x^2\sin x}{2}\]
+			\end{proof}
+			\item If $x^2 \cos x= \displaystyle\int_{0}^{x} f(t)\ dt$, find $f(x)$.
+			\\\\Since $\displaystyle\int_{0}^{x} f(t)\ dt = x^2\cos x$, we know that $x^2\cos x=F(x)$. Thus, in order to find $f(x)$, we must find $F'(x)=f(x)$. So,
+			\[\frac{d}{dx} x^2\cos x = 2x\cos(x)-x^2\sin(x)\]
+			And thus we have that $f(x)=2x\cos(x)-x^2\sin(x)$.\\
+			\item Let $F(x)=\displaystyle\int_{0}^{x} xe^{t^2}\ dt$ for $x \in [0,1]$. find $F''(x)$ for $x \in [0,1]$. (Note: $F'(x) \neq xe^{x^2}$) 
+			\begin{align*}
+				F(x) &= x \int_{0}^{x} e^{t^2}\ dt \\
+				F'(x) &= x \cdot e^{x^2} + \int_{0}^{x} e^{t^2}\ dt \\
+				F''(x) &= x \cdot e^{x^2} \cdot 2x + e^{x^2}+e^{x^2} \\
+				&= 2(x^2+1)e^{x^2}
+			\end{align*}
+			So $F''(x)=2(x^2+1)e^{x^2}$.\\
+		\end{enumerate}
+		\item Suppose $f$ is nonnegative and continuous on $[1,2]$ and that $\displaystyle\int_{1}^{2} x^k f(x)\ dx=5+k^2$ for $k=0,1,2$.
+		\\Prove each of the following:
+		\begin{enumerate}
+			\item $\displaystyle\int_{1}^{4} f(\sqrt{x})\ dx \leq 20$.\\\\
+			Let $u=\sqrt{x}$, then $du = u'dx = \frac{1}{2\sqrt{x}}\ dx$ and thus $dx = 2\sqrt{x}\ du$. So, 
+			\[\int_{1}^{4} f(\sqrt{x})\ dx = \int_{1}^{2} f(u)\ 2\ \sqrt{x}\ du = 2 \int_{1}^{2} \sqrt{x} f(u)\ du = 2 \int_{1}^{2} u f(u)\ du=2(5+1)=12\]
+			Thus $12 \leq 20$.\\
+			\item $\displaystyle\int_{1/\sqrt{2}}^{1} f(1/x^2)\ dx \leq 5/2$.\\\\
+			Substitute $u=\frac{1}{x^2}$ and $du=\frac{-2}{x^3}\ dx$. $\frac{1}{x^3}\ dx = \frac{-1}{2}\ du$.\\
+			We know $\frac{1}{\sqrt{2}} \leq x \leq 1$ and $\frac{1}{2\sqrt{2}} \leq x^3 \leq 1$. So $2\sqrt{2} \geq \frac{1}{x^3} \geq 1$. Thus $1 \leq \frac{1}{x^3}$ and $\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} 1 f \left(\frac{1}{x^2}\right) \leq \displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f\left(\frac{1}{x^2}\right)\ dx$. So
+			\[\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f \left(\frac{1}{x^2}\right)\ dx = \int_{2}^{1} \frac{-1}{2} f(u)\ du = \frac{1}{2} \int_{1}^{2} u^0 f(u)\ du = \frac{1}{2} \cdot 5 = \frac{5}{2}\]
+			Thus $\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{x^3} f \left(\frac{1}{x^2}\right)\ dx \leq \frac{5}{2}$.
+			\item $\displaystyle\int_{0}^{1} x^2 f(x+1)\ dx =2$. \\\\
+			Let $u=x+1$. Then $du=u'dx=1dx=dx$. Then,
+			\[\int_{0}^{1} x^2 f(x+1)\ dx = \int_{1}^{2} x^2 f(u)\ du\]
+			\begin{align*}
+				\int_{0}^{1} x^2 f(x+1)\ dx &= \int_{1}^{2} x^2 f(u)\ du \\
+				&= \int_{1}^{2} (u-1)^2f(u)\ du \\
+				&= \int_{1}^{2} (u^2-2u+1) f(u)\ du \\
+				&= \int_{1}^{2} u^2 f(u)\ du -2uf(u)+f(u)\ du \\
+				&= \int_{1}^{2} u^2f(u)\ du -2\int_{1}^{2} uf(u)\ du +\int_{1}^{2} f(u)\ du \\
+				&= (5+2^2)-2(5+1^2)+(5+0^2) \\
+				&= 9-12+5 \\
+				&= 2
+			\end{align*}
+			$\therefore\ \displaystyle\int_{0}^{1} x^2f(x+1)\ dx = 2$.\\
+		\end{enumerate}
+		\item Suppose that $f \in \mathcal{R}[1/2, 2]$ and that $\displaystyle\int_{1/2}^{1} x^k f(x)\ dx = \displaystyle\int_{1}^{2} x^kf(x)\ dx + 2k^2 = 3+k^2$ for $k=0,1,2$. Compute the exact values of the following integrals:
+		\begin{enumerate}
+			\item $\displaystyle\int_{0}^{1} x^3 f(x^2+1)\ dx$\\\\
+			Let $u=x^2+1$. Then $du=u'dx = 2x\ dx$. Thus $dx=\frac{du}{2x}$. Note that $x^2=u-1$. So, we have
+			\begin{align*}
+				\int_{0}^{1} x^3 f(x^2+1)\ dx &= \int_{0}^{1} x^3 f(u)\ \frac{du}{2x} \\
+				&= \int_{1}^{2} x^2 f(u)\ \frac{du}{2} \\
+				&= \frac{1}{2}\ \int_{1}^{2} (u-1) f(u)\ du \\
+				&= \frac{1}{2}\ \int_{1}^{2} uf(u)-f(u)\ du \\
+				&= \frac{1}{2}\ \int_{1}^{2} uf(u)\ du - \frac{1}{2}\ \int_{1}^{2} f(u)\ du \\
+				&= \frac{1}{2}\ \int_{1}^{2} uf(u)\ du - \frac{1}{2} \cdot (3+0) \\
+				&= \frac{1}{2} \left[\int_{1}^{2} uf(u)\ du + 3\right] \\
+				&= \frac{1}{2} \left[\int_{1}^{2} u f(u)\ du + 2 - 5\right] \\
+				&= \frac{1}{2} [3+1-5] \\
+				&= -\frac{1}{2}
+			\end{align*}
+			$\therefore\ \displaystyle\int_{0}^{1} x^3f(x^2+1)\ dx = -\frac{1}{2}$.\\
+			\item $\displaystyle\int_{0}^{\sqrt{3}/2} \frac{x^3}{\sqrt{1-x^2}} f(\sqrt{1-x^2})\ dx$\\\\
+			Let $u=\sqrt{1-x^2}$. Then $du=u'dx = \frac{-2x}{2\sqrt{1-x^2}}\ dx$. Thus
+			$dx=\frac{-2\sqrt{1-x^2}}{2x}\ du$. So,
+			\begin{align*}
+				\int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^3}{\sqrt{1-x^2}}f(\sqrt{1-x^2})\ dx &= \int_{1}^{\frac{1}{2}} \frac{x^3}{\cancel{u}} f(u) \cdot \frac{-2\cancel{u}}{2x}\ du \\
+				&= \int_{1}^{\frac{1}{2}} \frac{-x^3 f(u)}{x}\ du \\
+				&= \int_{1}^{\frac{1}{2}} -x^2f(u)\ du \\
+				&= -\int_{\frac{1}{2}}^{1} (u^2-1) f(u)\ du \\
+				&= \int_{\frac{1}{2}}^{1} (1-u^2) f(u)\ du \\
+				&= \int_{\frac{1}{2}}^{1} f(u)\ du - \int_{\frac{1}{2}}^{1} u^2 f(u)\ du \\
+				&= (3+0^2)-(3+2^2) \\
+				&= 3-7 \\
+				&= -4
+			\end{align*}
+			$\therefore\ \displaystyle\int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^3}{\sqrt{1-x^2}} f(\sqrt{1-x^2})\ dx = -4$.
+		\end{enumerate}
+		\item Suppose that $f,g$ are differentiable on $[0,e]$ and that $f', g' \in \mathcal{R}[0,e]$.
+		\begin{enumerate}
+			\item If $\displaystyle\int_{1}^{e} \frac{f(x)}{x}\ dx < f(e)$, prove that $\displaystyle\int_{1}^{e} f'(x) \ln x\ dx > 0$.
+			\\\[\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = \left.f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx\]
+			where $u=\ln (x)$, $dv=f'(x)\ dx$, $du=\frac{1}{x}\ dx$, $v=f(x)$. So 
+			\[\left.\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx=f(e)-\int_{1}^{e} \frac{1}{x}\ dx > 0\]
+			Since $\int_{1}^{e} \frac{f(x)}{x}\ dx < f(e)$.\\
+			\item If $f(0)=f(1)=0$, prove that $\displaystyle\int_{0}^{1} e^x [f(x)+f'(x)]\ dx =0$.
+			\begin{proof}
+				\[\int_{0}^{1}e^x \left[f(x)+f'(x)\right]\ dx = \int_{0}^{1} e^xf(x)\ dx +\int_{0}^{1} e^xf'(x)\ dx\]
+				Let us use \textit{Integration by Parts} on the second integral containing $f'(x)$. Let $u=e^x$. Then $du=e^x\ dx$, $dv=f'(x)\ dx$, and $v=f(x)$. Then we have the following:
+				\begin{align*}
+					\int_{0}^{1} e^xf(x)\ dx + \int_{0}^{1} f'(x)\ dx &= \left.\cancel{\int_{0}^{1} e^xf(x)\ dx} + e^xf(x)\right|_0^1 - \cancel{\int_{0}^{1} f(x)e^x\ dx} \\
+					&= e^1f(1)-e^0f(0) \\
+					&= e\cdot 0 - 1 \cdot 0 \\
+					&= 0 - 0 \\
+					&= 0
+				\end{align*}
+				$\therefore\ \displaystyle\int_{0}^{1} e^x\left[f(x)+f'(x)\right]\ dx = 0$.
+			\end{proof}
+		\end{enumerate}
+		\item
+		\begin{enumerate}
+			\item Let $f:[0,b] \to \R,\ b>0$ be continuous and $f(x) \neq 0$ for all $x \in (0,b)$. Further, suppose $[f(x)]^2 = 2 \displaystyle\int_{0}^{x} f(t)\ dt$ for all $x \in [0,b]$. Prove that $f(x)=x$ for all $x \in [0,b]$.\\\\We have
+			\[2f(x)f'(x)=2f(x)\]
+			which implies $f(x)[f'(x)-1]=0$. Since $f(x) \neq 0$, then $f'(x)-1=0$. So $f'(x)=1$ and $f(x)=x+C$ for some arbitrary constant $C$. But $f(0)=0$ since $[f(0)]^2=0\implies f(0)=0$. So $f(x)=x$.\\
+			\item Suppose that $f$ is defined on $[0,1]$ with $f(0)=0$ and $0<f'(x) \leq 1$. Prove that $\left[\displaystyle\int_{0}^{1} f(x)\ dx\right]^2 \geq \displaystyle\int_{0}^{1} [f(x)]^3\ dx$. 
+			\begin{proof}
+				Let $x \in [0,1]$. Then $F(x):=\big[\int_{0}^{x} f\big] ^2 - \int_{0}^{3} f^3$. So $f(0)=0$. Thus $F'(x)=2\left[\int_{0}^{x} f\right] \cdot f(x) - f^3(x)=f(x)\left[2\int_{0}^{x} f-f^2\right]$ since $f(0)=0$ and $0 \leq f'(x) \leq 1$ which implies that $f$ is strictly increasing. So $f(x)\geq 0$.
+			\end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 5.tex

@@ -1,7 +1,10 @@
 \documentclass[12pt,letterpaper]{article}
 \usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
 \usepackage[english]{babel}
 \usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
 \usepackage{amsmath}
 \usepackage{amsfonts}
 \usepackage{amssymb}
@@ -9,6 +12,8 @@
 \usepackage{array}
 \usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
 \usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
 \newcommand{\st}{\ \text{s.t.}\ }
 \newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
 \newcommand{\R}{\mathbb{R}}
@@ -16,419 +21,347 @@
 \newcommand{\Q}{\mathbb{Q}}
 \newcommand{\C}{\mathbb{C}}
 \newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
 \DeclareMathOperator{\sign}{sgn}
 \newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
 \theoremstyle{case}
 \newtheorem{case}{Case}
+\newtheorem{case*}{Case}
 \theoremstyle{definition}
 \newtheorem{definition}{Definition}[section]
 \newtheorem{theorem}{Theorem}[section]
 \newtheorem*{theorem*}{Theorem}
 \newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
 \newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
 \newtheorem*{remark}{Remark}
 \setlist[enumerate]{font=\bfseries}
 \renewcommand{\qedsymbol}{$\blacksquare$}
 \author{Alexander J. Tusa}
-\title{Real Analysis Homework 5}
+\title{Real Analysis II Homework 5}
 \begin{document}
 	\maketitle
 	\begin{enumerate}
-	\item For the following sequences, i) write out the first 5 terms, ii) Use the Monotone Sequence Property to show that the sequences converges.
+		\item Find the sum of the following series.
 		\begin{enumerate}
-		\item \textbf{Section 3.3}
+			\item (pr. 3a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2+3n+2}$
-		\begin{enumerate}
-			\item[2)] Let $x_1 > 1$ and $x_{n+1} := 2-1/x_n$ for $n \in \N$. Show that $(x_n)$ is bounded and monotone. Find the limit.
-			\\\\ The first five terms of this sequence are $x_1 \geq 2,x_2 \geq \frac{3}{2}, x_3 \geq \frac{4}{3}, x_4 \geq \frac{5}{4}, x_5 \geq \frac{6}{5}, \dots \approx x_1 \ge 2, x_2 \geq 1.5, x_3 \geq 1.3333, x_4 \geq 1.25, x_5 \geq 1.2, \dots$. This sequence appears to be decreasing.
-			\\\\Recall the Monotone Sequence Property:
-			\begin{theorem*}{Monotone Sequence Property}
-				A monotone sequence of real numbers is convergent if and only if it is bounded. Further,
-				\begin{enumerate}
-					\item If $X=(x_n)$ is a bounded increasing sequence, then
-					\[\lim (x_n) = \sup \{x_n:n \in \N\}\]
-					
-					\item If $Y=(y_n)$ is a bounded decreasing sequence, then
-					\[\lim (y_n) = \inf \{y_n : n \in \N \}\]
-				\end{enumerate}
-			\end{theorem*}
-		
-			To show that this sequence converges, we must first find the possible limit points (fixed points) of this sequence. So,
 			\begin{align*}
-				x&=2-\frac{1}{x} \\
+				\sum_{n=1}^{\infty} \frac{1}{n^2+3n+2} &= \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} \\
-				x^2 &= 2x -1 \\
+				&\Downarrow \\
-				x^2 - 2x + 1 &= 0 \\
+				\frac{1}{n^2+3n+2}&=\frac{A}{n+1} + \frac{B}{n+2} \\
-				(x-1)^2 &= 0
+				1&=A(n+2)+ B(n+2) \\
+				1&=An+2A+Bn+2B \\
+				1&=An+Bn+2A+2B \\
+				1&= (A+B)n+(A+B)2 \\
 			\end{align*}
-			Thus, $x=1$ is a possible limit of this sequence.
+			\[\begin{cases}
-			\\\\Now, we will prove that $(x_n)$ is bounded by $1$, and since we hypothesized that $(x_n)$ is decreasing, we say that $(x_n)$ is bounded below by 1. 
+			0=A+B \\
-			\begin{proof}
+			1=A+B
-				We want to show that the sequence $(x_n)$ is bounded below by 1; that is, we want to show that $1 \leq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
+			\end{cases} \implies
-				\\\\\textbf{Basis Step:} Let $n=1$. Then
+			\begin{cases}
+			A=1 \\
+			B=-1
+			\end{cases}\]
 			\begin{align*}
-					x_n &\geq x_{n+1}, &\text{by the definition of decreasing,} \\
+				&=\sum_{n=1}^{\infty} \frac{1}{n+1}-\frac{1}{n+2} \\
-					x_1 &\geq x_{1+1} \\
+				&= \left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \dots + \left(\cancel{\frac{1}{n}}-\cancel{\frac{1}{n+1}}\right)+ \left(\cancel{\frac{1}{n+1}}-\frac{1}{n+2}\right) \\
-					x_1 &\geq x_2
+				&= \limx{n}{\infty} \frac{1}{2}+\frac{1}{n+2} \\
+				&= \frac{1}{2}
 			\end{align*}
-				Since $x_1>1 \Rightarrow \frac{1}{x_1} < 1$, we have
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)}$
-					\[x_2 = 2-\frac{1}{x_1} > 1\]
-					\[\Rightarrow 1 < x_2 < 2.\]
-				Since $x_1>1$ and because $1 < x_2 < 2$, we have that $x_1 \geq x_2$.
-				\\\\\textbf{Inductive Step:} Assume $1 < x_n < 2,\ \forall\ n \in \N$.
-				\\\\\textbf{Show:} Now we want to show that $x_n \leq x_{n+1}$.
-				\\So,
-					\[1 < x_n <2\]
-					\[1 > \frac{1}{x_n} > \frac{1}{2}\]
-					\[-1 < -\frac{1}{x_n} < -\frac{1}{2}\]
-					\[1 < 2-\frac{1}{x_n} < 2-\frac{1}{2} < 2\]
-					\[1 < x_{n+1} < 2\]
-				Thus we have that $(x_n)$ is bounded between 1 and 2.
-			\end{proof}
-		
-			Now we need to show that $(x_n)$ is monotone decreasing; that is, we must show that $x_1 \geq x_2 \geq \dots \geq x_n$.
-			
-			\begin{proof}
-				We want to show that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
-				\\\\\textbf{Basis Step:} Let $n=1$. Then since $x_1 >1$ is given, we have that $\frac{1}{x_1} < 1$. This yields $x_2=2-\frac{1}{x_1}>1$, as was determined for the boundedness proof, and thus we have that $1 < x_2 < 2$. This means that $1 > \frac{1}{x_2} > \frac{1}{2}$, and since $\frac{1}{2} \leq \frac{1}{x_n}$, we have $x_2 \geq x_1$.
-				\\\\\textbf{Inductive Step:} Assume $x_n \geq x_{n+1}\ \forall\ n \in \N$.
-				\\\\\textbf{Show:} We now want to show that $x_{n+2} \leq x_{n+1}$.
-				\\So,
-				\[x_{n+2}=2-\frac{1}{x_{x+1}}\]
-				Recall the inductive hypothesis, in that $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$. Thus,
-				\[-\frac{1}{x_n} \geq -\frac{1}{x_{n+1}}\]
-				\[\Rightarrow 2-\frac{1}{x_n} \leq 2 -\frac{1}{x_{n+1}}\]
-				\[x_{n+1} \leq x_{n+2}\]
-				$\therefore$ we have that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$.
-			\end{proof}
-			Thus $(x_n)$ is monotone decreasing.
-			\\\\By the \textit{Monotone Sequence Property}, since we have shown that $(x_n)$ is both bounded (and thus converges), and that $(x_n)$ is monotone decreasing, we have that
 			\begin{align*}
-				\lim (x_n) &= \inf \{x_n: n \in \N\} \\
+				\frac{1}{(3n-2)(3n+1)} &= \frac{A}{3n-2} + \frac{B}{3n+1} \\
-				&=\inf (1,2) \\
+				1&= A(3n+1)+B(3n-2) \\
-				&= 1
+				1&= 3An+3Bn+A-2B
 			\end{align*}
-			Hence the sequence converges to the previously found possible limit of 1. \\
+			\[\begin{cases}
-			
+			3A+3B=0 \\
-			\item[3)] Let $x_1 > 1$ and $x_{n+1} := 1 + \sqrt{x_n - 1}$ for $n \in \N$. Show that $(x_n)$ is decreasing and bounded below by $2$. Find the limit.
+			1A-2B=1
-			\\\\The first 5 terms of this sequence are $x_1 \geq 2, x_2 \geq 2, x_3 \geq 2, x_4 \geq 2, x_5 \geq 2, \dots$. Notice the following, however:
+			\end{cases} \implies \begin{cases}
+			A=\frac{1}{3} \\
+			B=\frac{-1}{3}
+			\end{cases}\]
 			\begin{align*}
-				x_{n+1} \leq x_n &\iff 1+\sqrt{x_n - 1} \leq x_n \\
+				\frac{1}{(3n-2)(3n+1)} &= \frac{1}{9n-6} - \frac{1}{9n+3} \\
-				&\iff \sqrt{x_n -1} \leq x_n -1
+				\sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)} &= \sum_{n=1}^{\infty} \frac{1}{9n-6} - \frac{1}{9n+3} \\
+				&= \left(\frac{1}{3}-\cancel{\frac{1}{12}}\right) + \left(\cancel{\frac{1}{12}}-\cancel{\frac{1}{21}}\right) + \dots\\
+				&+ \left(\cancel{\frac{1}{9n-15}}-\cancel{\frac{1}{9n-6}}\right) + \left(\cancel{\frac{1}{9n-6}}-\frac{1}{9n+3}\right) \\
+				&= \limx{n}{\infty} \frac{1}{3}-\frac{1}{9n+3} \\
+				&= \frac{1}{3}
 			\end{align*}
-			which we know is always true since the square root function is a decreasing function.
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{e^{n-1}}$
-			\\\\Now we must find the possible limit points (fixed points) of this sequence. So,
 			\begin{align*}
-				x &= 1 + \sqrt{x-1} \\
+				\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{e^{n-1}} &= \sum_{n=1}^{\infty} \frac{(-1)^n \cdot (-1)^1}{e^n\cdot e^{-1}} \\
-				x-1 &= \sqrt{x-1} \\
+				&=- \sum_{n=1}^{\infty} \frac{(-1)^n \cdot e}{e^n} \\
-				x-1 &= (x-1)^2 \\
+				&= - \sum_{n=1}^{\infty} (-1)^n \cdot \frac{e}{e^n} \\
-				x-1 &= x^2 -2x +1 \\
+				&= - \sum_{n=1}^{\infty} (-1)^n \cdot e^{1-n} \\
-				(x-1)-(x^2-2x+1) &= 0 \\
+				&= \frac{e}{1+e}
-				-x^2+3x-2 &=0 \\
-				-(x^2-3x+2) &= 0 \\
-				-(x-1)(x-2) &= 0 \\
-				(x-1)(x-2) &= 0
 			\end{align*}
-			Thus $x=1$, or $x=2$. These are the possible limits of $(x_n)$. Since we hypothesized that $(x_n)$ is decreasing, then we say that $(x_n)$ is bounded below by 2, since we are given that $x_1 > 1$.
+			\item $\displaystyle\sum_{n=2}^{\infty} \frac{4^{n+1}}{9^{n-1}}$
-			\\\\Now we will prove that $(x_n)$ is bounded below by 2.\\
-			\begin{proof}
-				We want to show that $(x_n)$ is bounded below by 1; that is, we want to show that $1 \leq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
-				\\\\\textbf{Basis Step:} Let $n=1$. Then we are given that $x_1 \geq 2$.
-				\\\\\textbf{Inductive Step:} Assume that $x_n \geq 2,\ \forall\ n \in \N$.
-				\\\\\textbf{Show:} We now want to show that $x_{n+1} \geq 2,\ \forall\ n \in \N$.
-				\\\\So,
 			\begin{align*}
-					x_{n+1} &= 1+\sqrt{x_n -1} \\
+				\sum_{n=2}^{\infty} \frac{4^{n+1}}{9^{n-1}} &= \sum_{n=2}^{\infty} \frac{4^n \cdot 4^1}{9^n\cdot 9^{-1}} \\
-					&\geq 1+\sqrt{2 -1} \\
+				&= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 4^1 \cdot 9^1 \\
-					&=1 + 1 \\
+				&= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 36 \\
+				&= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 36 - 16-36 \\
+				&= a\left(\frac{1}{1-r}\right) \\
+				&= 36 \cdot \left(\frac{1}{1-\left(\frac{4}{9}\right)}\right) -52 \\
+				&= \frac{36 \cdot 9}{5} - 52 \\
+				&= \frac{324}{5} - 52 \\
+				&= \frac{324}{5} - \frac{260}{5} \\
+				&= \frac{64}{5}
+			\end{align*}
+			\item $\displaystyle\sum_{n=0}^{\infty} \frac{5^{n+1}+(-3)^n}{7^{n+2}}$
+			\begin{align*}
+				\sum_{n=0}^{\infty} \frac{5^{n+1}+(-3)^n}{7^{n+2}} &= \sum_{n=0}^{\infty} \frac{5^{n+1}}{7^{n+2}}+\frac{(-3)^n}{7^{n+2}} \\
+				&= \sum_{n=0}^{\infty} \frac{5}{49} \cdot \left(\frac{5}{7}\right)^n + \frac{1}{49} \cdot \left(\frac{(-3)}{7}\right)^n \\
+				&= \sum_{n=0}^{\infty} \frac{5}{49} \cdot \left(\frac{5}{7}\right)^n + \sum_{n=0}^{\infty} \frac{1}{49} \cdot \left(\frac{(-3)}{7}\right)^n \\
+				&= \frac{5}{49} \cdot \frac{1}{1-\frac{5}{7}} + \frac{1}{49} \cdot \frac{1}{1-\frac{-3}{7}} \\
+				&= \frac{5}{14} + \frac{1}{70} \\
+				&= \frac{13}{35}
+			\end{align*}
+			\item $\displaystyle\sum_{n=2}^{\infty} \ln \frac{n^2-1}{n^2}$
+			\begin{align*}
+				\sum_{n=2}^{\infty} \ln \left(\frac{n^2-1}{n^2}\right) &= \sum_{n=2}^{\infty} \ln \left(\frac{(n-1)(n+1)}{n^2}\right) \\
+				&= \sum_{n=2}^{\infty} \ln \left(\frac{\frac{n-1}{n}}{\frac{n}{n+1}}\right) \\
+				&= \sum_{n=2}^{\infty} \ln \left(\frac{n-1}{n}\right) - \ln\left(\frac{n}{n+1}\right) \\
+				&= \left(\ln \frac{1}{2} - \cancel{\ln \frac{2}{3}}\right) + \left(\cancel{\ln \frac{2}{3}} - \cancel{\ln \frac{3}{4}}\right) +\\
+				&\dots + \left(\cancel{\ln \frac{n-2}{n-1}}-\cancel{\ln \frac{n-1}{n}}\right) + \left(\cancel{\ln \frac{n-1}{n}}-\ln \frac{n}{n+1}\right) \\
+				&= \limx{n}{\infty} \ln \left(\frac{1}{2}\right) - \ln \left(\frac{n}{n+1}\right) \\
+				&= \ln \left(\frac{1}{2}\right) - \limx{n}{\infty} \frac{n+1}{n} \cdot \frac{n+1-n}{(n+1)^2}, &\text{by L'Hospital's Rule} \\
+				&= \ln \left(\frac{1}{2}\right) - \limx{n}{\infty} \frac{1}{n(n+1)} \\
+				&= \ln\left(\frac{1}{2}\right) - 0 \\
+				&= \ln\left(\frac{1}{2}\right) \\
+				&\approx -0.693147
+			\end{align*}
+			\item $\displaystyle\sum_{n=2}^{\infty} \ln \frac{n(n+2)}{(n+1)^2}$
+			\begin{align*}
+				\sum_{n=2}^{\infty} \ln\left(\frac{n(n+2)}{(n+1)^2}\right) &= \sum_{n=2}^{\infty} \ln \left(\frac{\frac{n}{n+1}}{\frac{n+1}{n+2}}\right) \\
+				&= \sum_{n=2}^{\infty} \ln\left(\frac{n}{n+1}\right) - \ln\left(\frac{n+1}{n+2}\right)  \\
+				&= \left(\ln \frac{2}{3}-\cancel{\ln \frac{3}{4}}\right) + \left(\cancel{\ln \frac{3}{4}}-\cancel{\ln \frac{4}{5}}\right) + \dots \\
+				&+ \left(\cancel{\ln \frac{n-1}{n}}-\cancel{\ln\frac{n}{n+1}}\right) + \left(\cancel{\ln \frac{n}{n+1}}-\ln\frac{n+1}{n+2}\right) \\
+				&= \limx{n}{\infty}\ln \left(\frac{2}{3}\right) - \ln \left(\frac{n+1}{n+2}\right) \\
+				&= \ln \left(\frac{2}{3}\right)-\limx{n}{\infty} \frac{n+2}{n+1} \cdot \frac{1\cdot (n+2) - (n+1)\cdot 1}{(n+2)^2}, &\text{by L'Hospital's Rule} \\
+				&= \ln \left(\frac{2}{3}\right)- \limx{n}{\infty} \frac{1}{(n+1)(n+2)} \\
+				&= \ln \left(\frac{2}{3}\right) - 0 \\
+				&= \ln \left(\frac{2}{3}\right) \\
+				&\approx -0.405465
+			\end{align*}
+			\item (pr. 3c) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$
+			\begin{align*}
+				\frac{1}{n(n+1)(n+2)} &= \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} \\
+				1&=A(n+1)(n+2) + Bn(n+2)+Cn(n+1) \\
+				1&=An^2+3An+2A+Bn^2+2Bn+Cn^2+Cn \\
+				1&= An^2+Bn^2+Cn^2+3An+2Bn+Cn+2A
+			\end{align*}
+			\[\begin{cases}
+			An^2+Bn^2+Cn^2=0 \\
+			3An+2Bn+Cn=0 \\
+			2A=1
+			\end{cases} = \begin{cases}
+			A=\frac{1}{2} \\
+			B=-1 \\
+			C=\frac{1}{2}
+			\end{cases}\]
+			\begin{align*}
+				\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} &= \sum_{n=1}^{\infty} \frac{1}{2n} -\frac{1}{n+1} + \frac{1}{2n+4} \\
+				&= \left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right) + \left(\frac{1}{4}-\frac{1}{3}+\frac{1}{8}\right) + \left(\frac{1}{6}-\frac{1}{4}+\frac{1}{10}\right) \\
+				&+ \dots + \left(\frac{1}{2n-2}+\frac{1}{2n+2}-\frac{1}{n}\right) + \left(\frac{1}{2n}+\frac{1}{2n+4}-\frac{1}{n+1}\right) \\
+				&= \limx{n}{\infty} \frac{1}{4} + \frac{1}{2(n+1)(n+2)} \\
+				&= \frac{1}{4} + 0 \\
+				&= \frac{1}{4}
+			\end{align*}
+			\item $\displaystyle\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots$
+			\\\\Notice that this is equal to the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$. So,
+			\begin{align*}
+				\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} &= \sum_{n=1}^{\infty} \frac{A}{2n-1} + \frac{B}{2n+1} \\
+				&\Downarrow \\
+				1 &= 2An+2Bn+A-B \\
+				\begin{cases}
+				2An+2Bn=0 \\
+				A-B=1
+				\end{cases} 
+				&\implies
+				\begin{cases}
+				A= \frac{1}{2} \\
+				B= \frac{-1}{2}
+				\end{cases} \\
+				&= \sum_{n=1}^{\infty} \frac{1}{4n-2} - \frac{1}{4n+2} \\
+				&= \left(\frac{1}{2}-\cancel{\frac{1}{6}}\right) + \left(\cancel{\frac{1}{6}}-\cancel{\frac{1}{10}}\right) + \\
+				&\dots + \left(\cancel{\frac{1}{4n-5}}-\cancel{\frac{1}{4n-2}}\right) + \left(\cancel{\frac{1}{4n-2}}-\frac{1}{4n+2}\right) \\
+				&=\limx{n}{\infty} \frac{1}{2} -\frac{1}{4n+2} \\
+				&= \frac{1}{2} - 0 \\
+				&= \frac{1}{2}
+			\end{align*}
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+2+3+\dots+n}$
+			\begin{align*}
+				\sum_{n=1}^{\infty} \frac{1}{1+2+3+\dots+n} &= \sum_{n=1}^{\infty} \frac{1}{\sum\limits_{i=1}^{n} i} \\
+				&= \sum_{n=1}^{\infty} \frac{2}{n(n+1)} \\
+				&\Downarrow \\
+				\frac{2}{n(n+1)} &= \frac{A}{n} + \frac{B}{n+1} \\
+				2&=An+A+Bn
+			\end{align*}
+			\[\begin{cases}
+			An+Bn=0 \\
+			Bn=2
+			\end{cases}=\begin{cases}
+			A=2 \\
+			B=-2
+			\end{cases}\]
+			\begin{align*}
+				\sum_{n=1}^{\infty} \frac{2}{n(n+1)} &= \sum_{n=1}^{\infty} \frac{2}{n}-\frac{2}{n+1} \\
+				&= \left(\frac{2}{1}-\cancel{\frac{2}{2}}\right) + \left(\cancel{\frac{2}{2}}-\cancel{\frac{2}{3}}\right) + \left(\cancel{\frac{2}{3}}-\cancel{\frac{2}{4}}\right) + \\
+				&\dots + \left(\cancel{\frac{2}{n-1}}-\cancel{\frac{2}{n}}\right) + \left(\cancel{\frac{2}{n}}-\frac{2}{n+1}\right) \\
+				&= \limx{n}{\infty} 2-\frac{2}{n+1} \\
+				&= 2-0 \\
 				&= 2
 			\end{align*}
-				Thus, $x_n \geq 2,\ \forall\ n \in \N$. By the definition of boundedness, we have that $(x_n)$ is bounded below by 2.
-			\end{proof}
-			
-			Since we have also shown earlier that $(x_n)$ is monotone decreasing, we have that by the monotone sequence property, since $(x_n)$ is bounded, $(x_n)$ converges, and since $(x_n)$ is monotone decreasing, we have:
-			\begin{align*}
-				\lim (x_n) &= \inf \{x_n:n \in \N\} \\
-				&=2
-			\end{align*}
-			
-			\item[7)] Let $x_1 := a>0$ and $x_{n+1} := x_n+1/x_n$ for $n \in \N$. Determine whether $(x_n)$ converges or diverges.
-			\\\\The first 5 terms of this sequence are $x_1 \geq 1, x_2 \geq 2, x_3 \geq \frac{5}{2}, x_4 \geq \frac{29}{10}, x_5 \geq \frac{941}{290}, \dots \approx x_1 \geq 1, x_2 \geq 2, x_3 \geq 2.5, x_4 \geq 2.9, x_5 \geq 3.244828, \dots$. This sequence appears to be increasing. We show this to be true as follows:
-			\begin{align*}
-				x_{n+1} \geq x_n &\iff x_n + \frac{1}{x_n} \geq x_n \\
-				&\iff x_n^2 + 1 \geq x_n^2 \\
-				&\iff 1 \geq 0
-			\end{align*}
-			which is true.
-			However, notice that one of the terms of the sequence is $x_n$. We know that $x_n$ is an unbounded sequence. Thus, we can infer that $(x_n)$ is unbounded above. We show this as follows:
-			\begin{align*}
-				x_{n+1}^2 &= \left(x_n + \frac{1}{x_n} \right)^2 \\
-				&= x_n^2+2+\frac{1}{x_n^2} \\
-				&> x_n^2 +2
-			\end{align*}
-			Since:
-			\[x_{n+1}^2 > x_n^2+2 > x_{n-1}^2 +4 > \dots > x_1^2+2 \cdot n = a^2+2 \cdot n\]
-			\[\Downarrow\]
-			\[x_n > \sqrt{a^2 + 2 \cdot (n-1)}\]
-			Since the right hand side of this inequality is unbounded, the left hand side is also unbounded.
-			\\\\Thus we have that this sequence $(x_n)$ is unbounded above.
-			\\\\Since this sequence is increasing and unbounded above, we have that the sequence is divergent.\\
-			
-			\item[8)] Let $(a_n)$ be an increasing sequence, $(b_n)$ be a decreasing sequence, and assume that $a_n \leq b_n$ for all $n \in \N$. Show that $\lim (a_n) \leq \lim (b_n)$, and thereby deduce the Nested Intervals Property 2.5.2 from the Monotone Convergence Theorem 3.3.2.
-			\\\\Since $(a_n)$ is an increasing sequence, we know that $(a_1 \leq a_2 \leq \dots \leq a_n)$, and since $(b_n)$ is a decreasing sequence, we know that $(b_1 \geq b_2 \geq \dots \geq b_n)$. Also, since we have that $a_n \leq b_n,\ \forall\ n \in \N$, we know that $(a_n)$ is bounded above by $(b_1)$. Thus, by the \textit{Monotone Convergence Theorem}, we know that
-			\[\lim (a_n) = \sup \{a_n: n \in \N\}\]
-			Also, since $(b_n)$ is a decreasing sequence such that it is bounded below by $(a_1)$, by the \textit{Monotone Convergence Theorem}, we have
-			\[\lim (b_n) = \inf \{b_n: n \in \N\}\]
-			Recall Theorem 3.2.5:
-			\begin{theorem*}
-				If $X=(x_n)$ and $Y=(y_n)$ are convergent sequences of real numbers and if $x_n \leq y_n\ \forall\ n \in \N$, then $\lim (x_n) \leq \lim (y_n)$.
-			\end{theorem*}
-			Also, recall the \textit{Nested Intervals Property}:
-			\begin{theorem*}
-				If $I_n=[a_n,b_n],\ n \in \N$, is a nested sequence of closed bounded intervals, then there exists a number $\xi \in \R \st \xi \in I_n\ \forall\ n \in \N$.
-			\end{theorem*}
-			Note that we have a nested sequence of closed, bounded intervals: $[a_n, b_n],\ n \in \N$. Since we showed that $\lim (a_n) \leq \lim (b_n)$, (and we are given that $(a_n)$ is increasing and $(b_n)$ is decreasing), we know that there exists $\xi$ such that
-			\[\lim (a_n) \leq \xi \leq \lim (b_n)\]
-			which means that $\xi \in [a_n, b_n],\ \forall\ n \in \N$.
 		\end{enumerate}
-		
+		\item Prove that each of the following series diverges.
-		\item $a_1 = 1,\ a_{n+1}=\frac{a_n^2+5}{2a_n}$
-		\\\\The first 5 terms of this sequence are $1, 3, \frac{7}{3}, \frac{47}{21}, \frac{2207}{987}, \dots \approx 1, 3, 2.3333, 2.2381, 2.2361, \dots$. This is a decreasing sequence.
-		\\\\First, we must find the possible limits (fixed points) of the sequence. So,
-		\begin{align*}
-			a&=\frac{a^2+5}{2a} \\
-			2a^2 &= a^2+5 \\
-			a^2 &= 5 \\
-			a &= \pm \sqrt{5}
-		\end{align*}
-		Since we're given that $a_1=1$, we know that the most likely lower bound will be $\sqrt{5}$.
-		\\\\Now we want to show that $(a_n)$ is bounded below by $\sqrt{5}$.
-		\begin{proof}
-			We want to show that $a_n \geq \sqrt{5},\ \forall\ n \in \N$. We prove it by method of mathematical induction.
-			\\\\\textbf{Basis Step:} Since $1 \geq \sqrt{5}$, we have that $a_1 \geq \sqrt{5}$
-			\\\\\textbf{Inductive Step:} Assume that $a_n \geq \sqrt{5}\ \forall\ n \in \N$.
-			\\\\\textbf{Show:} We want to show that $a_{n+1} \geq \sqrt{5}\ \forall\ n \in \N$. So, 
-			\[a_{n+1} = \frac{a_n^2 + 5}{2a_n}\]
-			\begin{align*}
-				(a_n-\sqrt{5})^2 &\geq 0 \\
-				a_n^2 -2\sqrt{5}a_n +5 &\geq 0 \\
-				a_n^2 +5 &\geq 2\sqrt{5}a_n \\
-				\Downarrow \\
-				\frac{a_n^2+5}{2a_n} &\geq \frac{2\sqrt{5}a_n}{2a_n} \\
-				\frac{a_n^2+5}{2a_n} &\geq \sqrt{5} \\
-				a_{n+1} \geq \sqrt{5}
-			\end{align*}
-			Thus we have that $(a_n)$ is bounded below by $\sqrt{5}$.
-		\end{proof}
-		Now we must show that $(a_n)$ is monotone decreasing.\\
-		\begin{proof}
-			We want to show that $(a_n)$ is monotone decreasing; that is, we want to show that $(a_2 \geq a_3 \geq \dots \geq a_n),\ \forall\ n \geq 2$. We prove it by method of mathematical induction.
-			\\\\\textbf{Basis Step:} Since $3 \geq \frac{7}{3}$, we have that $a_2 \geq a_3$.
-			\\\\\textbf{Inductive Step:} Assume that $a_n \geq a_{n+1},\ \forall\ n \geq 2$.
-			\\\\\textbf{Show:} We want to show that $a_{n+2} \leq a_{n+1},\ \forall\ n \geq 2$.
-			\\So,
-			\[a_{n+2} = \frac{a_{n+1}^2+5}{2a_{n+1}} \leq \frac{a_n^2+5}{2a_n}\]
-			Since we have:
-			\begin{align*}
-				a_{n+1} &\geq \sqrt{5}, &\text{by the previous proof of boundedness} \\
-				a_{n+1}^2 &\geq 5
-			\end{align*}
-			We can equivalently write the inequality as
-			\[\frac{a_{n+1}^2+5}{2a_{n+1}} \leq \frac{a_{n+1}^2+a_{n+1}^2}{2a_{n+1}}=a_{n+1}\]
-			Thus we have that $(a_n)$ is monotone decreasing.
-		\end{proof}
-		Since $(a_n)$ is both monotone decreasing and bounded, we have
-		\begin{align*}
-			\lim (a_n) &= \inf \{a_n:n \in \N\} \\
-			&= \sqrt{5}
-		\end{align*} 
-		
-		\item $a_1 = 5,\ a_{n+1}=\sqrt{4+a_n}$
-		\\\\The first 5 terms of this sequence are 5, 3, $\sqrt{7}$, $ \frac{\sqrt{14}}{2} + \frac{\sqrt{2}}{2} $, $ \frac{\sqrt{2 \cdot (\sqrt{14}+\sqrt{2}+8)}}{2} $, $\dots$, $\approx$ 5, 3, 2.64575131106, 2.57793547457, 2.5647486182, $\dots$. This sequence is decreasing.
-		\\\\First, we must find the possible limits (fixed points) of the sequence. So,
-		\begin{align*}
-			a&=\sqrt{4+a} \\
-			\sqrt{4+a} &= a \\
-			4+a &= a^2 \\
-			-a^2+a+4 &= 0 \\
-			a^2-a-4 &= 0 \\
-			a^2-a&=4 \\
-			a^2-a+\frac{1}{4}&=4+\frac{1}{4} \\
-			a^2-a+\frac{1}{4}&=\frac{17}{4} \\
-			(a-\frac{1}{2})^2&=\frac{17}{4} \\
-			a-\frac{1}{2}&=\pm \frac{\sqrt{17}}{2}
-		\end{align*}
-		So we have that $a=\frac{1}{2} + \frac{\sqrt{17}}{2}$, or $a=\frac{1}{2}-\frac{\sqrt{17}}{2}$. We must now check these solutions for correctness; so, 
-		\begin{align*}
-			a \Rightarrow \frac{1}{2}-\frac{\sqrt{17}}{2} &=\frac{1}{2} \left(1-\sqrt{17}\right) \\
-			&\approx -1.56155 \\\\
-			\sqrt{a+4} &= \sqrt{\left(\frac{1}{2}-\frac{\sqrt{17}}{2}\right)+4} \\
-			&=\frac{\sqrt{9-\sqrt{17}}}{\sqrt{2}} \\
-			&\approx 1.56155
-		\end{align*}
-		Thus, this solution is incorrect. Now we must validate that $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is correct. So,
-		\begin{align*}
-			a \Rightarrow \frac{1}{2}+\frac{\sqrt{17}}{2} &= \frac{1}{2}\left(1+\sqrt{17}\right) \\
-			&\approx 2.56155 \\\\
-			\sqrt{a+4} &= \sqrt{\left(\frac{\sqrt{17}}{2}+\frac{1}{2}\right)+4} \\
-			&= \frac{\sqrt{9+\sqrt{17}}}{\sqrt{2}} \\
-			&\approx 2.56155
-		\end{align*}
-		Thus $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is a correct solution.
-		\\\\Now we want to show that $(a_n)$ is bounded below by $\frac{1}{2}+\sqrt{17}$.
-		\begin{proof}
-			We want to show that $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$, by the definition of a lower bound. We prove this by method of mathematical induction.
-			\\\\\textbf{Basis Step:} Since $5 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$, we have that $a_1 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$.
-			\\\\\textbf{Inductive Step:} Assume $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$.
-			\\\\\textbf{Show:} We now want to show that $ a_{n+1} \geq \frac{1}{2}+\frac{\sqrt{17}}{2}\ \forall\ n \in \N$. So,
-			\begin{align*}
-				a_{n+1} &= \sqrt{4+a_n}, &\text{by the definition of the sequence} \\
-				&\geq \sqrt{4+\left(\frac{1}{2}+\frac{\sqrt{17}}{2}\right)}, &\text{by the inductive hypothesis} \\
-				&\geq \sqrt{\frac{8}{2}+\frac{1}{2}+\frac{\sqrt{17}}{2}} \\
-				&\geq \sqrt{\frac{9+\sqrt{17}}{2}} \\
-				&\geq \sqrt{\frac{1}{2}\left(9+\sqrt{17}\right)} \\
-				&\geq \sqrt{\frac{1}{4}+\frac{\sqrt{17}}{2}+\frac{17}{4}}, &\text{by expressing } \frac{9+\sqrt{17}}{2} \text{ as a square} \\
-				&\geq \sqrt{\frac{1+2\sqrt{17}+17}{4}} \\
-				&\geq \sqrt{\frac{1+2\sqrt{17}+(\sqrt{17})^2}{4}} \\
-				&\geq \sqrt{\frac{(\sqrt{17}+1)^2}{4}} \\
-				&\geq \sqrt{\frac{1}{4}\left(1+\sqrt{17}\right)^2} \\
-				&\geq \frac{\sqrt{(1+\sqrt{17})^2}}{\sqrt{4}} \\
-				&\geq \frac{\sqrt{17}+1}{2} \\
-				&\geq \frac{1}{2} + \frac{\sqrt{17}}{2}
-			\end{align*}
-			Thus we have that $ a_{n+1} \geq \frac{1}{2} + \frac{\sqrt{17}}{2}\ \forall\ n \in \N$.
-		\end{proof}
-		Now, we want to show that $ (a_n) $ is monotone decreasing; that is, we want to show that $(a_1 \geq a_2 \geq \dots \geq a_n)$.
-		\begin{proof}
-			We want to show that $(a_1 \geq a_2 \geq \dots \geq a_n),\ \forall\ n \in \N$. We prove this by method of mathematical induction.
-			\\\\\textbf{Basis Step:} Since $5 \geq 3$, we have that $a_1 \geq a_2$.
-			\\\\\textbf{Inductive Step:} Assume $a_n \geq a_{n+1}\ \forall\ n \in \N$.
-			\\\\\textbf{Show:} We want to show that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$. So,
-			\begin{align*}
-				a_{n+2} &= \sqrt{4+a_{n+1}} &\text{by the definition of the sequence} \\
-				&\leq \sqrt{4+a_n} &\text{by the inductive hypothesis} \\
-				&=a_{n+1}
-			\end{align*}
-			Thus we have that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$.
-		\end{proof}
-	Since $(a_n)$ is both bounded and monotone decreasing, by the \textit{Monotone Convergence Theorem}, we have that $(a_n)$ converges. Also by the \textit{Monotone Sequence Property}, we have that $(a_n)$ converges to the following:
-	\begin{align*}
-		\lim (a_n) &= \inf \{a_n: n \in \N\} \\
-		&= \frac{1}{2}+\frac{\sqrt{17}}{2} \approx 2.56155281281
-	\end{align*}
-	\end{enumerate}
-	
-	\item 
 		\begin{enumerate}
-		\item Show $a_n=\frac{3 \cdot 5 \cdot 7 \cdot \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$ converges to $A$ where $0 \leq A < 1/2$.
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n+1}$
-		\\\\\textbf{TODO}
-		
-		\item Show $b_n = \frac{2 \cdot 4 \cdot 6 \dots (2n)}{3 \cdot 5 \cdot 7 \dots (2n+1)}$ converges to $B$ where $0 \leq B < 2/3$.
-		\\\\\textbf{TODO}
-	\end{enumerate}
-
-	\item \textbf{Section 3.4}
-	\begin{enumerate}
-		\item[1)] Give an example of an unbounded sequence that has a convergent subsequence.
-		\\\\Let $a_n:=(-1)^n$. This sequence diverges since it oscillates between 1 and $-1$, however if we define $b_n:=(a_{2n})$; that is, let $(b_n)$ be the sequence of all terms of $(a_n)$ such that $n$ is even. Thus, $(b_n)$ is a subsequence of $(a_n)$, but $(b_n)$ converges since $a_2=1, a_4=1, \dots, a_{2n}=1$. Thus while $(a_n)$ is a divergent sequence, the subsequence $(b_n)$ of $(a_n)$ converges for all $n \in \N$.\\
-		
-		\item[3)] Let $(f_n)$ be the Fibonacci sequence of Example 3.1.2(d), and let $x_n := f_{n+1}/f_n$. Given that $\lim (x_n) =L$ exists, determine the value of $L$.
-		\\\\We can rewrite $(x_n)$ as follows:
-		\begin{align*}
-			x_n &= \frac{f_{n+1}}{f_n} \\
-			&=\frac{f_n+f_{n-1}}{f_n} \\
-			&= 1+\frac{f_{n-1}}{f_n} \\
-			&= 1+\frac{\frac{1}{f_n}}{f_{n-1}} \\
-			&= 1+\frac{1}{x_{n-1}}
-		\end{align*}
-		Since we're given that $L=\lim (x_n)$ exists and since we just showed that it's equal to $\lim (x_{n-1})$, we get the following:
-		\begin{align*}
-			x_n &= \left. 1+\frac{1}{x_{n-1}}\ \ \ \ \ \right| \lim \\
-			\lim (x_n) &= 1+\frac{1}{\lim(x_{n-1})} \\
-			L &= \left. 1+\frac{1}{L}\ \ \ \ \ \right| \cdot L \\
-			L^2 &= L + 1 \\
-			L^2-L-1 &= 0 \\
-			L_{1,2} &= \frac{1 \pm \sqrt{1-4 \cdot 1 \cdot (-1)}}{2} \\
-			L_1 &= \frac{1-\sqrt{5}}{2} <0 \\
-			L_2 &= \frac{1+\sqrt{5}}{2}>0
-		\end{align*}
-		Now, since $f_n >0 \Rightarrow x_n >0 \Rightarrow L>0$, we can infer that the proper limit is
-		\[L=\frac{1+\sqrt{5}}{2}\]
-		\\
-		
-		\item[4a)] Show that the sequence $(1-(-1)^n+1/n)$ converges.
-		\\\\ Let $(x_n):=(1-(-1)^n+1/n)$. Let $(z_n)=(x_{2n})$, and $(w_n)=(x_{2n-1})$ be subsequence of $(x_n)$. Then $(z_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is even, and $(w_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is odd.
-		\\\\These subsequences yield the following:
-		\[z_n = x_{2n} = 1-(-1)^{2n}+\frac{1}{2n}=1-1+\frac{1}{2n}=\frac{1}{2n}\]
-		\[w_n=x_{2n-1}=1-(-1)^{2n-1}+\frac{1}{2n-1}=1+1+\frac{1}{2n-1}=2+\frac{1}{2n-1}\]
-		Now, if we take the limit of each sequence as $n \rightarrow \infty$ yields
-		\[\lim_{n \to \infty} (z_n) = 0 \neq 2 = \lim_{n \to \infty} (w_n)\]
-		Recall Theorem 3.4.5 \textit{Divergence Criteria}:
-		\begin{theorem*}
-			If a sequence $X=(x_n)$ of real numbers has either of the following properties, then $X$ is divergent.
-			\begin{enumerate}
-				\item $X$ has two convergent subsequences $X'=(x_{n_k})$ and $X''=(x_{r_k})$ whose limits are not equal.
-				
-				\item $X$ is unbounded
-			\end{enumerate}
-		\end{theorem*}
-		Thus by the \textit{Divergence Criteria}, we have that since $(z_n)$ and $(w_n)$ satisfy the first property of the \textit{Divergence Criteria}, we can conclude that the sequence $(x_n)$ is divergent. \\
-		
-		\item[16)] Give an example to show that Theorem 3.4.9 fails if the hypothesis that $X$ is a bounded sequences is dropped.
-		\\\\Recall \textit{Theorem 3.4.9}:
-		\begin{theorem*}
-			If $(x_n)$ is a bounded sequence of real numbers, then the following statements for a real number $x^*$ are equivalent:
-			\begin{enumerate}
-				\item $x^*= \lim \sup (x_n)$.
-				
-				\item If $\varepsilon > 0$, there are at most a finite number of $n \in \N$ such that $x^*+\varepsilon<x_n$, but an infinite number of $n \in \N$ such that $x^*-\varepsilon<x_n$.
-				
-				\item If $u_m=\sup \{x_n : n \geq m\}$, then $x^*=\inf \{u_m:m \in \N\}=\lim (u_m)$.
-				
-				\item If $S$ is the set of subsequential limits of $(x_n)$, then $x^*= \sup (S)$.\\
-			\end{enumerate}
-		\end{theorem*}
-		Consider the sequence $((-1)^nn)$. We note that any subsequence of this sequence is unbounded and thus this sequence has no convergent subsequence. Due to this, all of the conditions of \textit{Theorem 3.4.9} are satisfied vacuously, save the condition concerning boundedness. However, this sequence doesn't converge, but both oscillates and diverges towards $\infty$ and $-\infty$. Thus if the boundedness criterion of the theorem is dropped, this theorem fails.
-		
-		\item[18)] Show that if $(x_n)$ is a bounded sequence, then $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
 			\begin{proof}
-			Let $(x_n)$ be a bounded sequence. We want to show that $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
+				Recall \textit{Theorem 3.7.1 -- The $n^{\text{th}}$-Term Test}: 
-			\\\\\textbf{TODO}
+				\begin{theorem*}[\textbf{The $n$th Term Test}]
+					If the series $\sum x_n$ converges, then $\lim (x_n) = 0$.
+				\end{theorem*}
+				Let $a_n$ be the sequence whose terms are obtained by $a_n:=\frac{n}{2n+1}$, for $n \in \N$. Then we have 
+				\[\limx{n}{\infty} a_n = \limx{n}{\infty} \frac{n}{2n+1} = \limx{n}{\infty} \frac{1}{2} \text{ by L'Hospital's Rule} = \frac{1}{2} \neq 0\]
+				Thus since $\limx{n}{\infty} a_n \neq 0$, we know that by \textit{Theorem 3.7.1}, $\displaystyle\sum_{n=1}^{\infty}$ is divergent.
 			\end{proof}
 		
-		\item[19)] Show that if $(x_n)$ and $(y_n)$ are bounded sequences, then
+			\item $\displaystyle\sum_{n=1}^{\infty} \cos \frac{1}{n^2}$
-		\[\lim \sup (x_n + y_n) \leq \lim \sup (x_n) + \lim \sup (y_n).\]
+			\begin{proof}
-		Give an example in which the two sides are not equal.
+				Let $a_n$ be the sequence whose terms are obtained by $a_n:= \cos \left(\frac{1}{n^2}\right)$, for $n \in \N$. Then we have 
-		\\\\\textbf{TODO}
+				\[\limx{n}{\infty} a_n = \limx{n}{\infty} \cos \left(\frac{1}{n^2}\right) = \cos (0) = 1\]
+				By \textit{The $n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, the series $\displaystyle\sum_{n=1}^{\infty} \cos \left(\frac{1}{n^2}\right)$ is divergent.
+			\end{proof}
+		
+			\item $\displaystyle\sum_{n=1}^{\infty} n \sin \frac{1}{n}$
+			\begin{proof}
+				Let $a_n$ be the sequence whose terms are obtained by $a_n:=n\sin\left(\frac{1}{n}\right)$, for $n \in \N$. Then we have
+				\[\limx{n}{\infty} a_n = \limx{n}{\infty} n \sin \left(\frac{1}{n}\right)=\limx{n}{\infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=\limx{n}{\infty} \frac{\frac{-\cos\left(\frac{1}{n}\right)}{n^2}}{-\frac{1}{n^2}}=\limx{n}{\infty} \cos\left(\frac{1}{n}\right) = \cos(0) = 1\]
+				By using \textit{L'Hospital's Rule}. Thus by the \textit{$n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, the sum $\displaystyle\sum_{n=1}^{\infty} n\sin\left(\frac{1}{n}\right)$ is divergent.
+			\end{proof}
+			\item $\displaystyle\sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right)^n$
+			\begin{proof}
+				Let $a_n$ be the sequence whose terms are obtained by $a_n:=\left(1-\frac{1}{n}\right)^n$, for $n \in \N$. Then, we have
+				\begin{align*}
+					\limx{n}{\infty} a_n &= \limx{n}{\infty} \left(1-\frac{1}{n}\right)^n \\
+					&= \limx{n}{\infty} e^{\ln \left(1-\frac{1}{n}\right)^n} \\
+					&= \limx{n}{\infty} \exp\left\{\ln\left(1-\frac{1}{n}\right)^n\right\} \\
+					&= \limx{n}{\infty} \exp \left\{n\ln\left(1-\frac{1}{n}\right)\right\} \\
+					&= \limx{n}{\infty} \exp\left\{\frac{\ln\left(1-\frac{1}{n}\right)}{\frac{1}{n}}\right\} \\
+					&= \limx{n}{\infty} \exp \left\{\frac{\frac{1}{1-\frac{1}{n}}\cdot \frac{1}{n^2}}{-\frac{1}{n^2}}\right\} \\
+					&= \limx{n}{\infty} \exp \left\{\frac{\frac{1}{n^2-n}}{-\frac{1}{n^2}}\right\} \\
+					&= \limx{n}{\infty} \exp\left\{-\frac{n^2}{n^2-n}\right\} \\
+					&= \limx{n}{\infty} \exp \left\{-\frac{n}{n-1}\right\} \\
+					&= \limx{n}{\infty} \exp \left\{-\frac{1}{1-\frac{1}{n}}\right\} \\
+					&= \exp \left\{-\frac{1}{1-0}\right\} \\
+					&= \exp (-1) \\
+					&= e^{-1} \\
+					&= \frac{1}{e}
+				\end{align*}
+				By using \textit{L'Hospital's Rule}. Thus by the \textit{$n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, we have that the sum $\displaystyle\sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right)^n$ diverges.
+			\end{proof}
 		\end{enumerate}
 		\item 
 		\begin{enumerate}
-		\item Show that $x_n=e^{\sin (5n)}$ has a convergent subsequence.
+			\item Give an example of two series $\sum a_k$ and $\sum b_k$ that differ in the first five terms, yet converge to the same value.
-		
+			\\\\Consider $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$. This series converges to 2. Also notice that $1+2+3+4-10\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = 2$. Thus the first five terms are different but converge to the same value.\\
-		\item Give an example of a bounded sequence with three subsequences converging to three different numbers.
+			\item Give an example of two series $\sum a_k$ and $\sum b_k$ that differ in infinitely many terms, yet converge to the same value.
-		
+			\\\\Consider the sums
-		\item Give an example of a sequence $x_n$ with $\lim \sup x_n = 5$ and $\lim \sup x_n = -3$.
+			\[\sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n\ \text{ and }\ \sum_{n=2}^{\infty} \frac{1}{n(n+1)}\]
-		
+			Let $a_n$ and $b_n$ be the sequences whose terms are obtained by $a_n:=\frac{15}{32}\left(\frac{1}{16}\right)^n$, for $n =0,1,2,3,\dots$, and let $b_n:=\frac{1}{n(n+1)}$, for $n \in \N$. Then we have 
-		\item Let $\lim \sup x_n = 2$. True or False: if $n$ is sufficiently large, then $x_n > 1.99$.
+			\[a_n:=\left(\frac{15}{32}, \frac{15}{512}, \frac{15}{8192}, \frac{15}{131072}, \frac{15}{2097152}, \dots\right)\]
-		
+			and
-		\item Compute the infimum, supremum, limit infimum, and limit supremum for $a_n = 3 - (-1)^n - (-1)^n/n$. 
+			\[b_n:=\left(\frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30}, \frac{1}{42},\dots\right)\]
+			It is clear that since the numerator of each term of $a_n$ is always $15$, and that the numerator of $b_n$ is always $1$, these two sequences are different at every term and thus differ in infinitely many terms, and thus the terms of the sums $\sum a_n$ and $\sum b_n$ also differ in infinitely many terms. Thus the first five terms of $\sum a_n$ are
+			\[\frac{15}{32},\ \frac{255}{512},\ \frac{4095}{8192},\ \frac{65535}{131072},\ \frac{1048575}{2097152},\ \dots\]
+			and the first five terms of $\sum b_n$ are
+			\[\frac{1}{6},\ \frac{1}{4},\ \frac{3}{10},\ \frac{1}{3},\ \frac{5}{14},\ \dots\]
+			However, notice that $\sum a_n$ and $\sum b_n$ converge to the  same value:
+			\begin{align*}
+			\sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n &= \frac{\frac{15}{32}}{1-\frac{1}{16}} \\
+			&= \frac{\frac{15}{32}}{\frac{15}{16}} \\
+			&= \frac{15 \cdot 16}{15 \cdot 32} \\
+			&= \frac{240}{480} \\
+			&= \frac{1}{2}
+			\end{align*}
+			and
+			\begin{align*}
+			\sum_{n=2}^{\infty} \frac{1}{n(n+1)} &= \sum_{n=2}^{\infty} \frac{A}{n} + \frac{B}{n+1} \\
+			1&=An+Bn+A \\
+			\begin{cases}
+			An+Bn=0 \\
+			A=1
+			\end{cases} &\implies \begin{cases}
+			A=1 \\
+			B=-1
+			\end{cases} \\
+			&\Downarrow \\
+			\sum_{n=2}^{\infty} \frac{1}{n(n+1)}&= \sum_{n=2}^{\infty}\frac{1}{n}-\frac{1}{n+1} \\
+			&= \left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) + \\
+			&\dots + \left(\cancel{\frac{1}{n-1}}-\cancel{\frac{1}{n}}\right) + \left(\cancel{\frac{1}{n}}-\frac{1}{n+1}\right) \\
+			&= \limx{n}{\infty} \frac{1}{2} - \frac{1}{n+1} \\
+			&= \frac{1}{2}-0 \\
+			&= \frac{1}{2}
+			\end{align*}
+			Thus we have that
+			\[\sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n = \sum_{n=2}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2}\]
+			\item Give an example of two series $\sum a_k$ and $\sum b_k$ that converge to real numbers $A$ and $B$, respectively, but the series $\sum a_kb_k$ converges to a value different from $AB$.
+			\\\\Consider the series $\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$ and $\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n$. Then we have
+			\[\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n=\frac{1}{1-\frac{1}{2}}=\frac{2}{1}=2\]
+			and
+			\[\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}\]
+			Note that $2 \cdot \frac{3}{2}=\frac{6}{2} = 3$. But the series
+			\[\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n\left(\frac{1}{3}\right)^n = \sum_{n=0}^{\infty} \left(\frac{1}{2} \cdot \frac{1}{3}\right)^n = \sum_{n=0}^{\infty} \left(\frac{1}{6}\right)^n = \frac{1}{1-\frac{1}{6}}=\frac{6}{5}\]
+			Thus we have that the product of the sums, 3, is not equal to the sum of the products, $\frac{6}{5}$.\\
+			\item Give an example of a series that diverges and whose sequence of partial sums is bounded.
+			\\\\Consider an alternating series, $\displaystyle\sum_{n=1}^{\infty} (-1)^n$. Then, note that $S_1:=-1,\ S_2:=-1+1=0,\ S_3:=-1+1-1=-1, S_4:=-1+1-1+1=0,\dots$. Then we have that the sequence of partial sums is bounded below by $-1$ and is bounded above by $1$. However, since this is an alternating series, we know that by the \textit{Geometric Series Test}, since $|r| = |-1|=1 \nless 1$, this series is divergent.
 		\end{enumerate}
 	
 		\item Prove or justify, if true. Provide a counterexample, if false.
 		\begin{enumerate}
-		\item If $a_n$ and $b_n$ are strictly increasing, then $a_n + b_n$ is strictly increasing.
+			\item If $a_n$ is strictly decreasing and $\limx{n}{\infty} a_n=0$, then $\sum a_n$ converges.
+			\\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$. Then, we have that $\limx{n}{\infty} \frac{1}{n} =0$, however since this is a harmonic series, we know that it is divergent, thus $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ is divergent.\\
 			
-		\item If $a_n$ and $b_n$ are strictly increasing, then $a_n \cdot b_n$ is strictly increasing.
+			\item If $a_n \neq b_n$ for all $n \in \N$ and if $\sum (a_n+b_n)$ converges, then either $\sum a_n$ converges or $\sum b_n$ converges.
+			\\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} (-1)^n$ and $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}$. Then we have that $\displaystyle\sum_{n=1}^{\infty} (-1)^n = -1+1-1+1-1+\dots$, which is divergent by the \textit{Geometric Series Test}, and $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} = 1-1+1-1+1-1+\dots$, which is also divergent by the \textit{Geometric Series Test}. However, $\displaystyle\sum_{n=1}^{\infty} (-1)^n+(-1)^{n+1} = 0+0+0+0+\dots = 0$ and thus converges. However, $a_n \neq b_n$ for all $n \in \N$ since
+			$a_n:= -1,1,-1,1,-1,1,\dots$ and $b_n:=1,-1,1,-1,1,-1,\dots$. Thus for all $n \in \N$, either $a_n=-1 \neq 1 = b_n$, or $a_n=1\neq -1=b_n$. Thus we have that $\sum (a_n+b_n)$ converges but neither $\sum a_n$ nor $\sum b_n$ converge.\\
 			
-		\item If $a_n$ and $b_n$ are monotonic, then $a_n + b_n$ is monotonic.
+			\item Suppose $\sum (a_n+b_n)$ converges. Then $\sum a_n$ converges if and only if $\sum b_n$ converges. 
+			\\\\This is a true statement since if $\sum (a_n+b_n)$ converges, then both $a_n+b_n$ converges, as was covered in our notes.\\
 			
-		\item If $a_n$ and $b_n$ are monotonic, then $a_n \cdot b_n$ is monotonic.
+			\item If $\limx{n}{\infty} a_n=A$, then $\displaystyle\sum_{n=1}^{\infty} (a_n-a_{n+2})=a_1+a_2-2A$.
+			\begin{proof}
+				Notice that we can rewrite the sum $\displaystyle\sum_{n=1}^{\infty} (a_n-a_{n+2})$ as $\displaystyle\sum_{n=1}^{\infty} \left((a_n-a_{n+1})+(a_{n+1}-a_{n+2})\right)$. Now, we have that the $n$th partial sum yields a telescoping series:
+				\[S_n:=[(a_1-\cancel{a_2})+(a_2-\cancel{a_3})]+ [(\cancel{a_2}-\cancel{a_3})+(\cancel{a_3}-\cancel{a_4})] + \dots + [(\cancel{a_n}-a_{n+1})+(\cancel{a_{n+1}}-a_{n+2})]\]
+				So $S_n=a_1+a_2-a_{n+1}-a_{n+2}$, which yields $\limx{n}{\infty} =a_1+a_2-A-A=a_1+a_2-2A$.
+			\end{proof}
 			
-		\item If a monotone sequence is bounded, then it is convergent.
+			\item $\sum a_n$ converges if and only if $\limx{n}{\infty} a_n=0$.
+			\\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$. This is the harmonic series, which we know diverges. However, $\limx{n}{\infty} \frac{1}{n} = 0$. Thus the limit is equal to 0, but the series does not converge.\\
 			
-		\item If a bounded sequence is monotone, then it is convergent.
+			\item Changing the first few terms in a series may affect the value of the sum of the series.
+			\begin{proof}
+				Suppose $x_n \to x$. Then for all $\varepsilon >0$, there exists some $N \in \N$ such that if $n \geq N$, then $|x_n-x| < \varepsilon$. Now, suppose $x'_n$ is a sequence such that for $n \geq M$, then $x'_n=x_n$.
+				\\\\Let $\varepsilon>0$ be given. Then there exists some $N \in \N$ such that for all $n \geq N$, $|x_n-x|<\varepsilon$. Let $N'=\max (N,M)$. Then if $n \geq N'$, we have that $|x'_n-x|<\varepsilon$. Hence $x'_n \to x$.
+				\\\\Consider a convergent series $\sum x_n$. If we let $s_n=x_1+x_2+\dots +x_n$, then we have that $s_n \to s$.
+				\\\\Consider the series $\sum x'_n$, where for $n \geq M$, then $x'_n=x_n$. Let $s'_n=x'_1+x'_2+\dots+x'_n$. Note that for $n \geq M$, we have $s'_n-s'_{M-1}=x'_M+\dots+x'_n=x_M+\dots + x_n$, and thus $s'_n-s'_{M-1}=s_n-s_{M-1}\to s-s_{M-1}$. Hence $s'_n \to (s-s_{M-1}+s'_{M-1})$.
+			\end{proof}
 			
-		\item If a convergent sequence is monotone, then it is bounded.
+			\item Changing the first few terms in a series may affect whether or not the series converges.
+			\\\\This is a false statement. Consider the telescoping series used previously, given by $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(n+1)}$. We know this series converges to $\frac{1}{2}$. Consider changing the first few terms as follows: 
+			\[\left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) +\dots \to \left(1-\cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) + \dots\]
+			Now we have that this series converges to 1, not $\frac{1}{2}$. However, despite changing the first few terms of the series, we did not change whether or not it converges. This is because the first few terms of a series can only finitely affect the sum. Thus, if a series converges, a finite change to the terms will still create a finite sum. Likewise, if the series diverges, a finite change will not allow the series to converge to a finite sum.\\
 			
-		\item If a convergent sequence is bounded, then it is monotone.
+			\item If $\sum a_n$ converges and $\limx{n}{\infty} \frac{a_n}{b_n}=0$, then $\sum b_n$ converges.
+			\\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$, which we note is the harmonic series. Then we have that $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1-\frac{1}{2}} = 2$ since it is a geometric series. Thus $\sum a_n$ converges. Also, notice that
+			\[\limx{n}{\infty} \left(\frac{\left(\frac{1}{2}\right)^n}{\frac{1}{n}}\right)=0\]
+			However, since $\sum b_n = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series, we know that it is divergent, and thus if $\sum a_n$ converges and $\limx{n}{\infty} \frac{a_n}{b_n}=0$, $\sum b_n$ can still be divergent.
 		\end{enumerate}
 	\end{enumerate}
 \end{document}

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 6}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item \textbf{Section 3.7}
+		\begin{enumerate}
+			\item[6.]
+			\begin{enumerate}
+				 \item Calculate the value of $\displaystyle\sum_{n=2}^{\infty} \left(\frac{2}{7}\right)^n$. (Note the series starts at $n=2$)
+				 
+				 \begin{align*}
+				 	\sum_{n=2}^{\infty} \left(\frac{2}{7}\right)^n &= \sum_{n=1}^{\infty} \left(\frac{2}{7}\right)^{n+2} \\
+				 	&= \sum_{n=1}^{\infty} \left(\frac{2}{7}\right)^n-\frac{2}{7} - 1 \\
+				 	&= \frac{1}{1-\frac{2}{7}} -\frac{2}{7} -1\\
+				 	&= \frac{1}{\frac{5}{7}} - \frac{2}{7} -1\\
+				 	&= \frac{7}{5} - \frac{2}{7} -1\\
+				 	&= \frac{4}{35}
+				 \end{align*}
+				 
+				 \item Calculate the value of $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{2n}$. (Note the series starts at $n=1$)
+				 
+				 \begin{align*}
+				 	\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{2n} &= \sum_{n=1}^{\infty} \left(\frac{1}{9}\right)^n \\
+				 	&= \sum_{n=1}^{\infty}\left(\frac{1}{9}\right)^n-1 \\
+				 	&= \frac{1}{1-\frac{1}{9}}-1 \\
+				 	&= \frac{1}{\frac{8}{9}} - 1 \\
+				 	&= \frac{9}{8} - 1 \\
+				 	&= \frac{1}{8}
+				 \end{align*}
+			\end{enumerate}
+			\item[7.] Find a formula for the series $\displaystyle\sum_{n=1}^{\infty} r^{2n}$ when $|r| < 1$.
+			\\\\Let $S_n = r^2+r^4+r^6+\dots+r^{2(n-1)}+r^{2n}$. Then, we notice that $S_n=(r^2)^1+(r^2)^2+(r^2)^3+\dots + (r^2)^{n-1} + (r^2)^n$ which in turn yields $S_n=r^2[1+r^2+(r^2)^2+(r^2)^3+\dots+(r^2)^{n-1}]$ yielding $S_n=r^2 \cdot \frac{1-(r^2)^n}{1-r^2}=\frac{r^2}{1-r^2} \cdot (1-(r^2)^n)$. Since $|r|<1$, we know that $|r^2|<1$, and thus $\limx{n}{\infty} (r^2)^n=0$. So $\lim S_n=\frac{r^2}{1-r^2} \cdot (1-0)=\frac{r^2}{1-r^2}$. Thus $\displaystyle\sum_{n=1}^{\infty} r^{2n}$ converges and is equal to $\frac{r^2}{1-r^2}$.
+			\item[9.]
+			\begin{enumerate}
+				\item Show that the series $\displaystyle\sum_{n=1}^{\infty} \cos n$ is divergent.
+				\\\\We note that if the series converges, then $\limx{n}{\infty} \cos n = 0$. Consider the subsequence $n_k=2k\pi$, for some $k \in \N$. Then $\limx{k}{\infty} \cos (2k\pi)=1$. And for the subsequence $n_k=\frac{\pi}{2}+2k\pi$, for some $k \in \N$, then we have that $\limx{k}{\infty} \cos \left(\frac{\pi}{2}+2k\pi\right)=0$. And since $0 \neq 1$, we can conclude that $\cos n$ does not converge and that $\limx{n}{\infty} \cos n \neq 0$. Therefore, by the \textit{nth Term Test}, $\displaystyle\sum_{n=1}^{\infty} \cos n$ diverges.
+				 
+				\item Show that the series $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ is convergent.
+				\\\\We note that $\abs{\frac{\cos n}{n^2}} \leq \frac{1}{n^2}$, and since $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent $p$-series, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ is also convergent.
+			\end{enumerate}
+			\item[11.] If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum a^2_n$ always convergent? Either prove it or give a counterexample.
+			\begin{proof}
+				Since $\sum a_n$ is convergent, we know that by the \textit{nth Term Test}, $\lim a_n=0$. This yields that for $\varepsilon=1>0$, there exists $N \in \N$ such that for all $n \geq N$, we have
+				\begin{align*}
+					|a_n-0| &<\varepsilon \\
+					|a_n|&<\varepsilon \\
+					|a_n|&<1 
+				\end{align*}
+				\[0<a_n<1\]
+				since $a_n > 0,\ \forall n$. Since we know that $x^2<x$ for $0 < x < 1$, we have that $0<a_n^2<a_n<1,\ \forall n \geq N$. Thus, by the \textit{Comparison Test}, since $0 \leq a_n^2 \leq a_n$ and since $\sum a_n$ converges, we have that $\sum a_n^2$ converges.
+			\end{proof}
+			
+			\item[12.] If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum \sqrt{a_n}$ always convergent? Either prove it or give a counterexample.
+			\\\\This is a false statement. Consider $\sum a_n = \sum \frac{1}{n^2}$. Then we have that $\sum \sqrt{a_n} = \sum \frac{1}{n}$, which is a harmonic series, which we know diverges. Thus if $\sum a_n$ converges, then $\sum \sqrt{a_n}$ does not necessarily converge.\\
+			
+			\item[13.] If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum \sqrt{a_na_{n+1}}$ always convergent? Either prove it or give a counterexample.
+			\begin{proof}
+				We first notice that 
+				\[\frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}\]
+				So, we then have
+				\[\frac{a_1}{2}+\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)=\frac{a_1}{2}+\frac{a_1}{2}+\frac{a_2}{2}+\frac{a_2}{2} + \frac{a_3}{2} + \frac{a_3}{2} + \dots = a_1+a_2+a_3+\dots = \sum_{n=1}^{\infty}a_n\]
+				thus, since $\frac{a_1}{2}+ \displaystyle\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)=\displaystyle\sum_{n=1}^{\infty}a_n$, and since we're given that $\sum a_n$ converges, then $\displaystyle\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)$ converges. Thus by the \textit{Comparison Test}, we have that 
+				\[a_n > \left(\frac{a_n+a_{n+1}}{2}\right)>\sqrt{a_na_{n+1}}\]
+				yields that $\displaystyle\sum_{n=1}^{\infty} \sqrt{a_na_{n+1}}$ also converges.
+			\end{proof}
+			
+			\item[16.] Use the \textit{Cauchy Condensation Test} to discuss the $p$-series $\displaystyle\sum_{n=1}^{\infty} (1/n^p)$ for $p>0$.
+			\\\\By the \textit{Cauchy Condensation Test}, we must show that the series $\displaystyle\sum_{n=0}^{\infty}2^n\cdot a(2^n)$ converges. So,
+			\begin{align*}
+				\sum_{n=0}^{\infty} 2^na(2^n) ^= \sum_{n=0}^{\infty} \left[2^n\frac{1}{(2^n)^p}\right] \\
+				&= \sum_{n=0}^{\infty} 2^n \left[\frac{1}{2^{np}}\right] \\
+				&= \sum_{n=0}^{\infty} 2^n (2^{-np}) \\
+				&= \sum_{n=0}^{\infty} 2^{n(1-p)} \\
+				&= \sum_{n=0}^{\infty} (2^{1-p})^n
+			\end{align*}
+			We notice that this is now a geometric series, with $|r|=|2^{1-p}|$. Then we note that if $|2^{1-p}| \leq 1$, the series converges, and otherwise the series diverges.\\\\
+		
+			\item[17.] Use the \textit{Cauchy Condensation Test} to establish the divergence of the series:
+			\begin{enumerate}
+				\item $\displaystyle\sum \frac{1}{n \ln n}$
+				\\\\By the \textit{Cauchy Condensation Test}, we have
+				\begin{align*}
+					\sum \frac{1}{n \ln n} &= \sum 2^n \cdot \frac{1}{2^n \cdot \ln 2^n} \\
+					&= \sum \frac{1}{\ln 2^n} \\
+					&= \sum \frac{1}{n \ln 2} \\
+					&= \frac{1}{\ln 2} \sum \frac{1}{n}
+				\end{align*}
+				Since $\sum \frac{1}{n}$ is a harmonic series, we know that the series diverges. Thus we have that $\sum 2^n \cdot \frac{1}{2^n \cdot \ln 2^n}$ diverges as well. Thus by the \textit{Cauchy Condensation Test}, we have that $\sum \frac{1}{n \ln n}$ diverges also.\\
+				
+				\item $\displaystyle\sum \frac{1}{n(\ln n)(\ln\ln n)}$
+				\\\\By the \textit{Cauchy Condensation Test}, we have
+				\begin{align*}
+					\sum \frac{1}{n (\ln n)(\ln \ln n)} &= \sum 2^n \frac{1}{2^n\cdot (ln 2^n) \cdot (\ln (\ln 2^n))} \\
+					&= \sum \frac{1}{\ln 2^n \cdot \ln \ln 2^n} \\
+					&= \sum \frac{1}{(n \ln 2) \cdot (\ln (n \ln 2))} \\
+					&= \sum \frac{1}{(n \ln 2) \cdot (\ln n+\ln\ln 2)} \\
+					&> \sum \frac{1}{n \cdot (\ln 2 + \ln \ln 2} &(\ln 2 < 1) \\
+					&> \sum \frac{1}{n \cdot \ln n}
+				\end{align*}
+				Since we showed in part (a) that $\sum 2^n \cdot \frac{1}{n \ln n}$ diverges, and thus by the \textit{Comparison Test}, since $0 \leq \frac{1}{n(\ln n)(\ln\ln n)} < \frac{1}{n \ln n}$, since $\sum \frac{1}{n \ln n}$ diverges, then so does $\sum \frac{1}{(\ln 2^n)(\ln\ln 2^n)}$.\\
+			\end{enumerate}
+		\end{enumerate}
+	
+		\item Use the tests in section 3.7 to test the given series for convergence or divergence. State clearly which test is used. Also, for parts a-f, write out the first three terms of each series.
+		\begin{enumerate}
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{2n+5}{3n^2+2n-1}$
+			\\\\First three terms:
+			\[\sum_{n=1}^{\infty} \frac{2n+5}{3n^2+2n-1} = \frac{7}{4}+\frac{3}{5}+\frac{11}{32}+\dots\]
+			Note that $2n<2n+5$ and that $3n^2+2n-1<4n^2 \implies \frac{1}{3n^2+2n-1} > \frac{1}{4n^2}$, and thus $\frac{2n}{4n^2} = \frac{1}{2n} < \frac{2n+5}{3n^2+2n-1}$. Since $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}$ yields a half times the harmonic series, which we know is divergent, we have that by the \textit{Comparison Test}, $\sum \frac{2n+5}{3n^2+2n-1}$ also diverges.\\
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$
+			\\\\The first three terms:
+			\[\sum_{n=1}^{\infty} \frac{n-1}{n2^n} = 0+ \frac{1}{8} + \frac{1}{12}+\dots\]
+			We note that $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$ looks similar to $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$, which is a geometric series. We also note that the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges since $\abs{\frac{1}{2}} < 1$. Then, by the \textit{Limit Comparison Test}, we have $\limx{n}{\infty} \left(\frac{\frac{n-1}{n2^n}}{\left(\frac{1}{2}\right)^n}\right)=1 \neq 0$, and thus since $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges, then $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$ must also converge.\\
+			
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n} + \pi}{2+\sqrt[5]{n^8}}$
+			\\\\The first three terms of this series are:
+			\[\sum_{n=1}^{\infty} \frac{\sqrt{n} + \pi}{2+\sqrt[5]{n^8}} = \frac{\pi + 1}{3} + \frac{\pi + \sqrt{2}}{2+2 \sqrt[5]{8}}+\frac{\pi + \sqrt{3}}{2+3\sqrt[5]{27}}+ \dots\]
+			We note that $\sqrt{n}+\pi < \sqrt{n}$ and that $2+\sqrt[5]{n^8} > n^{\frac{7}{4}} \implies \frac{1}{2+\sqrt[5]{n^8}} < \frac{1}{n^{\frac{7}{4}}}$. Thus we have that
+			\[\frac{\sqrt{n}+\pi}{2+\sqrt[5]{n^8}}< \frac{\sqrt{n}}{n^{\frac{7}{4}}}=\frac{1}{n^\frac{5}{4}}\]
+			We note that $\sum \frac{1}{n^\frac{5}{4}}$ converges because it is a $p$-series where $p > 1$. Thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}+\pi}{2+\sqrt[5]{n^8}}$ must also converge.\\
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}$
+			\\\\The first three terms are:
+			\[\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}=1+\frac{1}{e^{(\ln 2)^2}}+\frac{1}{e^{(\ln 3)^2}}+\dots\]
+			We note that $\frac{1}{n^{\ln n}}$ looks like $\frac{1}{n^{\ln 3}}$ and that $0 < \displaystyle\frac{1}{n^{\ln n}} \leq \frac{1}{n^{\ln 3}}$. By the \textit{Comparison Test}, since $\displaystyle\frac{1}{n^{\ln 3}}$ is a convergent $p$-series since $\ln 3 \geq 1$, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}$ must also be convergent. \\
+			
+			\item $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$
+			\\\\The first three terms of the series are:
+			\[\sum_{n=2}^{\infty} \frac{1}{\ln n}=\frac{1}{\ln (2)} + \frac{1}{\ln (3)} + \frac{1}{\ln (4)} + \dots\]
+			We note that $n > \ln n \implies \frac{1}{n} < \frac{1}{\ln n}$. Then by the \textit{Comparison Test}, since $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n}$ is the Harmonic series, we know that it diverges, and thus by the \textit{Comparison Test}, since $\sum \frac{1}{n} \leq \frac{1}{\ln n}$, we have that $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$ must also diverge.\\
+			\item $\displaystyle\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2+1}$
+			\\\\The first three terms of the series are:
+			\[\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2+1}=\frac{\ln(2)+2}{5}+\frac{\ln(3)+3}{10} + \frac{\ln(4)+4}{17}+\dots\]
+			We notice that $\displaystyle\frac{n+\ln n}{n^2+1}$ looks like $\displaystyle\frac{1}{n}$. So, by the \textit{Limit Comparison Test}, we have
+			\begin{align*}
+				\limx{n}{\infty} \frac{\frac{n+\ln n}{n^2+1}}{\frac{1}{n}} &= \limx{n}{\infty} \frac{n^2+n\ln n}{n^2 +1} \\
+				&= \limx{n}{\infty} \frac{1+\frac{\ln n}{n}}{1+\frac{1}{n^2}} \\
+				&= \frac{1+0}{1+0} \\
+				&= \frac{1}{1} \\
+				&= 1
+			\end{align*}
+			Thus by the \textit{Limit Comparison Test}, since $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n}$ diverges, we have that $\displaystyle\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2 + 1}$ diverges. \\
+			
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^2}$
+			\\\\We note that $n < n+1$ and that $n^2 > \frac{n^2}{2} \implies \frac{1}{n^2}<\frac{2}{n^2}$ and gives us that
+			\[\frac{2n}{n^2}=\frac{2}{n}<\frac{n+1}{n^2}\]
+			Since $\displaystyle 2 \sum_{n=1}^{\infty} \frac{1}{n}$ is a harmonic series, it is divergent. Thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^2}$ is also divergent.\\
+			
+			\item $\displaystyle\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$
+			\\\\By the \textit{Cauchy Ratio Test}, we have the following:
+			\[\limx{n}{\infty} \abs{\frac{\frac{1}{\ln(n+1)^{n+1}}}{\frac{1}{\ln(n)^n}}}=\limx{n}{\infty} \abs{\frac{\ln(n)^n}{\ln(n+1)^{n+1}}}=0\]
+			Thus by the \textit{Cauchy Ratio Test}, since $L=0<1$, we have that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\ln (n)^n}$ is a convergent series. \\
+			
+			\item $\displaystyle\sum_{n=2}^{\infty} \frac{\ln n}{n^2}$
+			\\\\Notice that $\frac{\ln n}{n^2} < \frac{1}{n^2}$. So, we must note that $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ is a convergent series since it is a $p$-series with $p=2$ and $p>0$. Thus by the \textit{Comparison Test}, we have that the series $\displaystyle\sum_{n=2}^{\infty} \frac{\ln (n)}{n^2}$  is a convergent series.\\
+			
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$
+			\\\\We notice that $\frac{\ln n}{n}$ looks like $\frac{1}{n}$. So, by the \textit{Limit Comparison Test}, we have
+			\[\limx{n}{\infty}\frac{\frac{\ln n}{n}}{\frac{1}{n}}= \limx{n}{\infty} \ln n = \infty\]
+			Since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, we know that it diverges, and thus by the \textit{Limit Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ diverges as well.\\
+			
+			\item $\displaystyle\sum_{n=3}^{\infty} \frac{\sqrt{n}}{\sqrt{n^3}+1}$
+			\\\\Notice that $\sqrt{n^3} < 2\sqrt{n^3} \implies \frac{1}{\sqrt{n^3}+1} > \frac{1}{2\sqrt{n^3}}$, which gives us that
+			\[\frac{\sqrt{n}}{2\sqrt{n^3}}<\frac{\sqrt{n}}{\sqrt{n^3}+1}\]
+			Since the series $\displaystyle \frac{1}{2}\sum_{n=3}^{\infty} \frac{1}{n}$ is a harmonic series ,we know that it diverges, and thus we know that by the \textit{Comparison Test} we have that the series $\displaystyle\sum_{n=3}^{\infty} \frac{\sqrt{n}}{\sqrt{n^3}+1}$ is also divergent.\\
+			
+			\item $\displaystyle\sum_{n=1}^{\infty} \frac{n}{e^n}$
+			\\\\By the \textit{McClaurin Integral Test}, we have
+			\[\int_{1}^{\infty}\frac{x}{e^x}\ dx = \int_{1}^{\infty}xe^{-x}\ dx\]
+			By \textit{Integration by Parts}, let $u=x,\ dv=e^{-x}dx,\ du=dx,\ v=-e^{-x}$. Then we have
+			\begin{align*}
+				\int_{1}^{\infty} xe^{-x}\ dx &= -xe^{-x}+\int_{1}^{\infty} e^{-x}\ dx \\
+				&= \left.-xe^{-x}-e^{-x}\right|_1^\infty \\
+				&= \frac{-\infty}{e^\infty}-\frac{1}{e^\infty} +\frac{1}{e}+\frac{1}{e} \\
+				&= \frac{2}{e}
+			\end{align*}
+			Thus, since $\displaystyle\int_{1}^{\infty} \frac{x}{e^x}\ dx$ converges to $\frac{2}{e}$, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{n}{e^n}$ converges.
+		\end{enumerate}
+	
+		\item Write the given expressions as a quotient of two integers.
+		\begin{enumerate}
+			\item $3+1+\frac{1}{2} + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots $
+			\begin{align*}
+				&= 3+1+\frac{1}{2}+\sum_{n=1}^{\infty} \frac{1}{3^n} \\
+				&= 3+1+\frac{1}{2}+\frac{\frac{1}{3}}{1-\frac{1}{3}} \\
+				&= 3+1+\frac{1}{2}+\frac{\frac{1}{3}}{\frac{2}{3}} \\
+				&= 3+1+\frac{1}{2}+\frac{3}{6} \\
+				&= 3+1+\frac{1}{2}+\frac{1}{2} \\
+				&= 3+1+1 \\
+				&= 5
+			\end{align*}
+			
+			\item $3.2\overline{15}$
+			\begin{align*}
+				3.2\overline{15} &= 3+ \frac{2}{10} + \left(\frac{15}{1000}+\frac{15}{100000}+\frac{15}{10^7}+\dots\right) \\
+				&= 3+\frac{2}{10}+\frac{15}{1000}\left(1+\frac{1}{10^2}+\frac{1}{10^3}+\dots\right)\\
+				&= 3+\frac{2}{10} + \frac{\frac{15}{1000}}{1-\frac{1}{100}} \\
+				&= 3+\frac{2}{10} + \frac{1}{66} \\
+				&= \frac{1061}{330}
+			\end{align*}
+		\end{enumerate}
+		\item 
+		\begin{enumerate}
+			\item Let $a_n$ be a sequence of real numbers and $b_n=a_n-a_{n-1}$ for all $n \in \N$. Prove: $\sum b_k$ converges if and only if the sequence $a_n$ converges. In this case, find the sum of $\sum b_k$.
+			\begin{proof}
+				Let $(a_n)\subseteq \R$ and let $b_n:=a_n-a_{n-1}\ \forall\ n \in \N$.
+				\begin{itemize}
+					\item[$(=>)$] Assume $\sum b_k$ converges. Then by the \textit{Cauchy Criterion for Series}, we have
+					\[\forall\ \varepsilon>0,\ \exists\ M(\varepsilon)\in \N \st \text{if}\ m>n\geq M(\varepsilon) \implies |s_m-s_n|<\varepsilon\]
+					This yields that
+					\begin{align*}
+						|s_m-s_n|&= |(\cancel{a_{n+1}}-a_n)+(\cancel{a_{n+2}}-\cancel{a_{n+1}})+\dots+(\cancel{a_{m-1}}-\cancel{a_{m-2}})+(a_m-\cancel{a_{m-1}})| \\
+						&= |a_m-a_n| \\
+						&<\varepsilon
+					\end{align*}
+					Thus by the definition of a \textit{Cauchy Sequence}, $a_n$ is a Cauchy sequence and thus by the \textit{Cauchy Convergence Criterion}, $a_n$ is a convergent sequence.
+					
+					\item[$(<=)$] Similarly, suppose $a_n$ is a convergent sequence. Then by the \textit{Cauchy Convergence Criterion}, $a_n$ is a Cauchy sequence. So $\forall\ \varepsilon>0,\ \exists\ H(\varepsilon) \in \N \st \forall\ m>n\geq H(\varepsilon),\ n,m \in \N,\ |a_m-a_n|<\varepsilon$. So,
+					\begin{align*}
+						|a_m-a_n| &= |(a_{n+1}-a_n)+(a_{n+2}-a_{n+1})+\dots+(a_{m-1}-a_{m-2})+(a_m-a_{m-1})| \\
+						&= |(s_{m-1}+a_m)-(s_{n-1}+a_n)|, \\
+						&\text{by the definition of the infinite series generated by $(a_n)$}\\
+						&=|s_{m-1}+a_m-s_{n-1}-a_n|\\
+						&= |s_m-s_n| \\
+						&<\varepsilon
+					\end{align*}
+					Thus by the \textit{Cauchy Criterion for Series}, since $|s_m-s_n|<\varepsilon$ for all $m>n\geq H(\varepsilon), H(\varepsilon),m,n\in \N$, the series $\sum a_n-a_{n-1}$ converges, which implies that the series $\sum b_n$ converges.
+				\end{itemize}
+			\end{proof}
+			Thus, we have that $\sum b_n=\limx{n}{\infty} (b_n)=\limx{n}{\infty} (a_n-a_{n-1})$, by \textit{Theorem 3.7.3}, and thus by the \textit{$n$th Term Test}, since $\sum b_n$ converges, we have that $\limx{n}{\infty} (a_n-a_{n-1})=0$. Thus $\sum b_n=0$.\\
+			
+			\item Let $a_n$ be a sequence of real numbers. If $\limx{n}{\infty} a_n=A$, find the sum of \[\displaystyle\sum_{n=1}^{\infty} (a_{n+1}-2a_n+a_{n-1})\]
+			By the \textit{Cauchy Convergence Criterion for Series}, we have that $\forall\ \varepsilon>0,\ \exists\ M(\varepsilon) \in \N \st \text{if } m,n \in \N,\ m>n\geq M(\varepsilon) \implies |s_m-s_n|=|a_{n+1}+a_{n+2}+\dots + a_m|<\varepsilon$. So,
+			\begin{align*}
+				|s_m-s_n|&=|(\cancel{a_{n+2}}-\cancel{2}a_{n+1}+a_n)+(\cancel{a_{n+3}}-\cancel{2a_{n+2}}+\cancel{a_{n+1}}) \\
+				&+(\cancel{a_{n+4}}-\cancel{2a_{n+3}}+\cancel{a_{n+2}})+(\cancel{a_{n+5}}-\cancel{2a_{n+4}}+\cancel{a_{n+3}})\\
+				&+\dots\\
+				&+(\cancel{a_m}-\cancel{2a_{m-1}}+\cancel{a_{m-2}})+(a_{m+1}-\cancel{2}a_m+\cancel{a_{m-1}})| \\
+				&= |a_n-a_{n+1}+a_{m+1}-a_m| \\
+				&<\varepsilon
+			\end{align*}
+			This yields that the sequence of partial sums of $a_{n+1}-2a_n+a_{n-1}$ is bounded. Thus by \textit{Theorem 3.7.3}, $\displaystyle\sum_{n=1}^{\infty}(a_{n+1}-2a_n+a_{n-1})=\limx{n}{\infty} (a_{n+1}-2a_n+a_{n-1}) = A-2A+A = 0$. \\
+			
+			\item If $\displaystyle\sum_{k=1}^{n} (k\ a_k)=\frac{n+1}{n+2}$ for $n \in \N$, show that $\displaystyle\sum_{k=1}^{\infty} a_k=\frac{3}{4}$.
+			\begin{proof}
+				If $n=1$, then $\displaystyle\sum_{k=1}^{1} 1\cdot a_1 = \frac{2}{3}$. 
+				\\\\If $n \geq 2$, then
+				\begin{align*}
+					\sum_{k=1}^{n} ka_k &= a_1+a_2+\dots+(n-1)a_{n-1}+na_n \\
+					&= \frac{n+1}{n+2} \\
+					&\Downarrow \\
+					na_n &= \frac{n+1}{n+2} - \frac{n}{n+1} \\
+					&= \frac{1}{(n+1)(n+2)}
+				\end{align*}
+				So $a_n=\frac{1}{n(n+1)(n+2)}$, thus we have the following:
+				\begin{align*}
+					\sum_{n=2}^{\infty} &= \sum_{n=2}^{\infty} \frac{1}{n(n+1)(n+2)} \\
+					&= \frac{1}{2} \sum_{n=2}^{\infty} \left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right] \\
+					&= \frac{1}{12} &\text{as we did on the previous homework}
+				\end{align*}
+				So $\displaystyle\sum_{n=1}^{\infty} a_n=\frac{2}{3}+\frac{1}{12} = \frac{3}{4}$.
+			\end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 7}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item \textbf{Section 9.1}
+		\begin{enumerate}
+			\item[7.]
+			\begin{enumerate}
+				\item If $\sum a_n$ is absolutely convergent and $(b_n)$ is a bounded sequence, show that $\sum a_nb_n$ is absolutely convergent.
+				\begin{proof}
+					We want to show that $\sum (a_nb_n)$ is also absolutely convergent.\\
+					Since $(b_n)$ is bounded, we know that there is a $M > 0$ such that $|b_n| \leq M,\ \forall\ n$. Then we have that
+					\[|a_nb_n|=|a_n|\cdot|b_n|\leq M \cdot |a_n|\]
+					Since $\sum a_n$ is absolutely convergent, $M \cdot \sum |a_n|$ is also convergent. And since $|a_nb_n| \leq M \cdot |a_n|$ we know that $\sum |a_nb_n|$ is also convergent, and therefore $\sum (a_nb_n)$ is absolutely convergent.
+				\end{proof}
+			
+				\item Give an example to show that if the convergence of $\sum a_n$ is conditional and $(b_n)$ is a bounded sequence, then $\sum a_nb_n$ may diverge.
+				\\\\Consider the series $\sum a_n=\sum\displaystyle\frac{(-1)^n}{n}$ and the bounded sequence  $(b_n)=(-1)^n$. We know that $\sum a_n$ is conditionally convergent and $(b_n)$ is bounded by 1. And since $\sum (a_nb_n) = \sum \displaystyle\frac{(-1)^n}{n}\cdot(-1)^n=\sum \frac{1}{n}$, we have that the product series is a harmonic series, and thus diverges.\\\
+			\end{enumerate}
+			\item[8.] Give an example of a convergent series $\sum a_n$ such that $\sum a^2_n$ is not convergent. (Compare this with Exercise 3.7.11)
+			\\\\Consider the series $\sum a_n=\sum \displaystyle\frac{(-1)^n}{\sqrt{n}}$, which we know is convergent. But, $\sum (a_n)^2 = \sum \frac{1}{n}$, which is a harmonic series, and thus diverges.\\
+			\item[9.] If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum a_n$ is convergent, show that $\lim (na_n)=0$.
+			\begin{proof}
+				Let $(a_n)$ be a sequence such that $a_1 \geq a_2 \geq \dots \geq a_n \geq 0$, and let $\sum a_n$ be convergent, and let $s_n=a_1+a_2+\dots+a_n$.
+				\\\\Let $\varepsilon>0$ be given. Since $\sum a_n$ is convergent, we know that $\limx{n}{\infty} s_n=0$ by the \textit{$n$th-Term Test}. By the \textit{Cauchy Criterion for Series}, $\exists\ M(\varepsilon) \in \N \st$ if $m > n \geq M(\varepsilon)$, then $|s_m-s_n| = |a_{n+1}+a_{n+2}+\dots+a_m|<\varepsilon$. Now, since $(a_n)$ is a decreasing sequence, we have
+				\begin{align*}
+					s_{2n}-s_n &= (a_1+\dots+a_{2n}) - (a_1+\dots+a_n) \\
+					&= a_{n+1}+\dots + a_{2n} \\
+					&\geq a_{2n}+\dots+a_{2n} \\
+					&= n \cdot a_{2n}\\
+					&> 0
+				\end{align*}
+				Additionally, we note that
+				\begin{align*}
+					s_{2n-1}-s_n &= (a_1+\dots +a_{2n-1}) -(a_1+\dots+a_n) \\
+					&= a_{n+1}+\dots + a_{2n-1} \\
+					&\geq a_{2n-1}+\dots+a_{2n-1} \\
+					&=(n-1)\cdot a_{2n} \\
+					&> 0
+				\end{align*}
+				Notice that if we let $n>M(\varepsilon)$, then $|s_{2n}-s_n| < \varepsilon$ and $|s_{2n-1}-s_n| <\varepsilon$.
+				\\\\Choose $\delta = 2M(\varepsilon)$ and for $n > \delta$, we get $n\cdot a_{2n} \cdot (n-1)\cdot a_{2n} = na_n < \varepsilon$, and thus by the definition of a limit, we have that 
+				\[\limx{n}{\infty} na_n=0\]
+			\end{proof}
+		
+			\item[10.] Give an example of a divergent series $\sum a_n$  with $(a_n)$ decreasing and such that $\lim (na_n)=0$.
+			\\\\Consider the series $\sum a_n=\sum \frac{1}{n \ln n}$. We showed on the previous homework that this series diverges. And the sequence $(a_n)$ is decreasing since $n \ln n<(n+1)\ln(n+1)$. Thus, we have
+			\[\limx{n}{\infty} na_n = \limx{n}{\infty} n \cdot \frac{1}{n \ln n} = \limx{n}{\infty} \frac{1}{\ln n} = 0\]
+			
+			\item[11.] If $(a_n)$ is a sequence and if $\lim (n^2a_n)$ exists in $\R$, show that $\sum a_n$ is absolutely convergent.
+			\begin{proof}
+				Since $\lim na^2_n$ exists, we know that the sequence $(n^2a_n)$ is bounded. Thus we know that there exists $M>0$ such that 
+				\[|n^2a_n| \leq M \implies |a_n| \leq \frac{M}{n^2},\ \forall\ n\]
+				Since we know that $\sum \frac{M}{n^2}$ is convergent, by the \textit{Comparison Test}, we know that $\sum |a_n|$ is also convergent, which yields that $\sum a_n$ is absolutely convergent.
+			\end{proof}
+		
+			\item[12.] Let $a > 0$. Show that the series $\sum (1+a^n)^{-1}$ is divergent if $0<a\leq 1$ and is convergent if $a>1$.
+			\begin{proof}
+				Let $a > 0$. We want to show that the series $\displaystyle \sum a_n=\sum\frac{1}{1+a^n}$ is divergent when $0 < a \leq 1$ and is convergent when $a > 1$.
+				\begin{case}[$0<a<1$]
+					We notice that
+					\[\limx{n}{\infty} a_n = \limx{n}{\infty} \frac{1}{1+a^n}=\frac{1}{1+\limx{n}{\infty}a^n} = \frac{1}{1+0}=1 \neq 0\]
+					And thus by the \textit{$n$th Term Test}, we have that $\sum a_n$ is divergent when $0 < a < 1$.
+				\end{case}
+				\begin{case}[$a=1$]
+					We notice that
+					\[\limx{n}{\infty} a_n = \limx{n}{\infty} \frac{1}{1+1^n}=\limx{n}{\infty} \frac{1}{2} = \frac{1}{2} \neq 0\]
+					and thus by the \textit{$n$th Term Test}, the sum $\sum a_n$ is divergent when $a=1$.
+				\end{case}
+				\begin{case}[$a>1$]
+					We notice that 
+					\[\frac{1}{1+a^n}<\frac{1}{a^n}=\left(\frac{1}{a}\right)^n\]
+					which yields a geometric series, and since $a > 1 \implies \left(\frac{1}{a}\right) < 1$, by the \textit{Geometric Series Test}, the series $\sum \left(\frac{1}{a}\right)^n$ converges, and by the \textit{Comparison Test}, the series $\sum \frac{1}{1+a^n}$ must also converge.
+				\end{case}
+			\end{proof}
+			\item[13.]
+			\begin{enumerate}
+				\item Does the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right)$ converge?
+				\begin{align*}
+					\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right) &= \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}} \cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} \\
+					&= \sum_{n=1}^{\infty} \frac{(n+1)-n}{\sqrt{n}\cdot (\sqrt{n+1}+\sqrt{n})} \\
+					&= \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}\cdot (\sqrt{n+1}+\sqrt{n})} \\
+					&= \sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}+n}
+				\end{align*}
+				We notice that $\displaystyle\frac{1}{\sqrt{n(n+1)}+n} \approx \frac{1}{2n}$ and that
+				\begin{align*}
+					\limx{n}{\infty} \frac{\frac{1}{\sqrt{n(n+1)}+n}}{\frac{1}{2n}} &= \limx{n}{\infty} \frac{2n}{\sqrt{n(n+1)}+n} \\
+					&= \limx{n}{\infty} \frac{2n}{\sqrt{n(n+1)}+n} \cdot \frac{\frac{1}{n}}{\frac{1}{n}} \\
+					&= \limx{n}{\infty} \frac{2}{\sqrt{1(1+\frac{1}{n})+1}} \\
+					&= \frac{2}{\sqrt{1(1+0)}+1} \\
+					&= 1
+				\end{align*}
+				Thus we know that since $\sum \frac{1}{n}$ is a harmonic series, it diverges, and thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$ diverges.\\\\
+				
+				\item Does the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{n}\right)$ converge?
+				\begin{align*}
+					\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} &= \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} \cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} \\
+					&= \sum_{n=1}^{\infty} \frac{(n+1)-n}{n\cdot (\sqrt{n+1}+\sqrt{n})} \\
+					&= \sum_{n=1}^{\infty} \frac{1}{n\cdot(\sqrt{n+1}+\sqrt{n})}
+				\end{align*}
+				We notice that 
+				\[\frac{1}{n \cdot (\sqrt{n+1}+\sqrt{n})} \leq \frac{1}{n \cdot \sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}\]
+				And since we know that $\displaystyle\sum \frac{1}{n^{\frac{3}{2}}}$ is a convergent $p$-series, we have that by the \textit{Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$ must also be convergent.
+			\end{enumerate}
+		\end{enumerate}
+	
+		\item 
+		\begin{enumerate}
+			\item  If $a_n \geq 0$ for all $n \in \N$ and $\displaystyle\sum_{n=1}^{\infty} a_n$ converges, prove that $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n^p}$ converges for all $p \geq 0$.
+			\begin{proof}
+				We notice that the series $\displaystyle\sum \frac{a_n}{n^p}$ looks like $\displaystyle\sum \frac{1}{n^p}$, which is a convergent $p$-series when $p > 1$. So, by the \textit{Limit Comparison Test}, we have for $p >1$:
+					\[\limx{n}{\infty} \frac{\frac{a_n}{n^p}}{\frac{1}{n^p}}=\limx{n}{\infty} \frac{a_nn^p}{n^p} = \limx{n}{\infty} a_n = 0\] 
+				Since by the \textit{$n$th Term Test}, since $\sum a_n$ converges, $\limx{n}{\infty} a_n = 0$. And since $\sum \frac{1}{n^p}$ is a convergent $p$ series, by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{a_n}{n^p}$ is convergent when $p > 1$.
+				\\\\As for the case in which $0 \leq p \leq 1$, consider the following subcases:
+				\begin{case}[$p=0$]
+					\[\sum_{n=1}^{\infty} \frac{a_n}{n^p} = \sum_{n=1}^{\infty} \frac{a_n}{n^0} = \sum_{n=1}^{\infty} \frac{a_n}{1} = \sum_{n=1}^{\infty} a_n\]
+					Which since we assumed that $\sum a_n$ converges, we have that when $p=0$, $\sum \frac{a_n}{n^p}$ converges.
+				\end{case}
+				\begin{case}[$p=1$] We notice that $\frac{a_n}{n} \leq \frac{na_n}{n}$, and thus we have that by the \textit{Comparison Test}, $\sum \frac{na_n}{n} = \sum a_n$, which we know converges, and thus by the \textit{Comparison Test}, $\sum \frac{a_n}{n}$ must also converge.
+				\end{case}
+				\begin{case}[$0<p<1$]
+					We notice that $0\leq\frac{a_n}{n^p} < \frac{a_n\cdot \sqrt{n^p}}{\sqrt{n^p}}$. So, by the \textit{Comparison Test}, we have
+					\[\sum_{n=1}^{\infty} \frac{a_n \cdot \sqrt{n^p}}{\sqrt{n^p}} = \sum_{n=1}^{\infty} a_n\]
+					And since $\sum a_n$ converges, we have that by the \textit{Comparison Test}, $\sum \frac{a_n}{n^p}$ must also converge.
+				\end{case}
+			Therefore $\forall\ p \geq 0$, if $\sum a_n$ converges, then $\sum \frac{a_n}{n^p}$ converges.
+			\\\\Alternatively, we notice that $p \geq 0$ implies that $n^p \geq 1$. This yields that $0 \leq \frac{1}{n^p} \leq 1$. Then, since $\sum a_n$ converges, by the \textit{Comparison Test}, we have that $\sum \frac{a_n}{n^p}$ must also converge.
+			\end{proof}
+			
+			\item Let $a_k$ be a sequence of nonnegative real numbers. Prove: If $\sum a_n$ converges, then its sequence of partial sums is bounded.
+			\begin{proof}
+				Refer to \textit{Theorem 3.7.3}:
+				\begin{theorem*}
+					Let $(x_n)$ be a sequence of nonnegative real numbers. Then the series $\sum x_n$ converges if and only if the sequence $S=(s_k)$ of partial sums is bounded. In this case,
+					\[\sum_{n=1}^{\infty} x_n = \lim (x_n) = \sup \{s_k:k \in \N\}\]
+				\end{theorem*}
+			\end{proof}
+			\item Give an example of a series that diverges and whose sequence of partial sums is bounded.
+			\\\\Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ where $\frac{(-1)^n}{n}\neq 0$. This is an alternating series, whose sequence of partial sums is
+			\[(s_k)=\left(-1,-\frac{1}{2},-\frac{5}{6}, -\frac{7}{12}, \dots\right)\]
+			We have that this sequence is bounded below by -1, and is bounded above by 0, however this series diverges since by the \textit{Alternating Series Test}, $\limx{n}{\infty} \frac{1}{n} = \infty$, and thus the series diverges but the sequence of partial sums is bounded.\\
+			
+			\item Prove: If $\sum |a_n|$ converges and a sequence $b_n$ is bounded, then $\sum a_nb_n$ converges.
+			\begin{proof}
+				Refer to Problem 7a from Section 9.1:
+				
+				We want to show that $\sum (a_nb_n)$ is also absolutely convergent.\\
+				Since $(b_n)$ is bounded, we know that there is a $M > 0$ such that $|b_n| \leq M,\ \forall\ n$. Then we have that
+				\[|a_nb_n|=|a_n|\cdot|b_n|\leq M \cdot |a_n|\]
+				Since $\sum a_n$ is absolutely convergent, $M \cdot \sum |a_n|$ is also convergent. And since $|a_nb_n| \leq M \cdot |a_n|$ we know that $\sum |a_nb_n|$ is also convergent, and therefore $\sum (a_nb_n)$ is absolutely convergent.
+				\\\\By \textit{Theorem 9.1.1}, since $\sum |a_nb_n|$ is absolutely convergent, then the series $\sum a_nb_n$ is also convergent.
+			\end{proof}
+		\end{enumerate}
+	
+		\item (pr. 18a, Sec. 3.7) Find the positive values of $p$ such that the logarithmic $p$-series $\displaystyle\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^p}$ converges using a) the integral test and b) the Cauchy condensation test.
+		\\\\Since the terms are decreasing, we can use the \textit{Cauchy Condensation Test}.
+		\begin{align*}
+			\sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{\infty}\frac{1}{n(\ln n)^p} \\
+			\sum_{n=1}^{\infty} 2^na_{2n} &= \sum_{n=1}^{\infty} 2^n\cdot \frac{1}{2^n(\ln 2^n)^p} \\
+			&= \sum_{n=1}^{\infty} \frac{1}{(n \ln 2)^p} &(\ln a^b = b \ln a) \\
+			&= \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^p} \cdot \frac{1}{n^p} \\
+			&= \frac{1}{(\ln 2)^p} \sum_{n=1}^{\infty} \frac{1}{n^p}
+		\end{align*}
+		Now, since $\sum \frac{1}{n^p}$ is a $p$-series, the only way for the series to converge is if $p >1$ by the \textit{$p$-Series Test}, and thus we have that by the \textit{Cauchy Condensation Test} and by the \textit{Comparison Test}, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n (\ln n)^p}$ converges if and only if $p > 1$.
+		\\\\So, by the \textit{Integral Test}, let $u=\ln k$ and $du = k\ dk$. Then we have:
+		\[\int_{2}^{\infty} \frac{1}{k(\ln k)^p}\ dk = \int_{2}^{\infty} \frac{1}{u^p}\ du = \int_{\ln 2}^{\infty} u^{-p}\ du = \limx{b}{\infty} \left.\frac{u^{-p+1}}{1-p}\right|_{\ln 2}^b=\limx{b}{\infty} \frac{b^{-p+1}}{1-p}-\frac{(\ln 2)^{-p+1}}{1-p}\]
+		which we have converges only when $p > 1$ since only then will the $b$ term go to 0.
+		\\\\And by the \textit{Cauchy Condensation Test}, we have $\displaystyle\sum_{n=2}^{\infty} \frac{2^n}{2^n(\ln 2^n)^p}=\sum_{n=2}^{\infty} \frac{1}{(\ln 2^n)^p}=\sum_{n=2}^{\infty} \frac{1}{(n \ln 2)^p} = \frac{1}{(\ln 2)^p} \sum_{n=2}^{\infty} \frac{1}{n^p}$, and by the definition of a $p$-series this converges when $p>1$.
+		
+		\item Give an example of two series $\sum a_n$ and $\sum b_n$ that converge but $\sum a_nb_n$ diverges. (Similar to pr. 8, Sec. 9.1)
+		\\\\Consider the series $\sum a_n=\sum \displaystyle\frac{(-1)^n}{\sqrt{n}}$ and $\sum a_n=\sum \displaystyle\frac{(-1)^{n+1}}{\sqrt{n}}$ which we know is convergent. But, $\sum (a_nb_n) = \sum -\frac{1}{n}$, which is a negative harmonic series, and thus diverges.\\
+		
+		\item Prove or justify, if true; Provide a counterexample, if false.
+		\begin{enumerate}
+			\item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_n$ and $\sum b_n$ converge, then $\sum a_nb_n$ converges.
+			\\\\This is a true statement.
+			\begin{proof}
+				Since $\sum a_n$ converges, by the \textit{$n$th Term Test}, $\lim a_n = 0$. So there must exists some $N \in \N \st\ \forall\ n \geq N,\ a_n \leq 1$. So, $0 \leq a_n \leq 1$, which yields that $0 \leq a_nb_n \leq b_n\ \forall\ n \geq N$. Thus, since $\sum b_n$ converges, by the \textit{Comparison Test}, we have that $\sum a_nb_n$ converges.
+			\end{proof}
+			
+			\item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_nb_n$ converges, then $\sum a_n$ and $\sum b_n$ converge. 
+			\\\\This is a false statement. Consider the series $\sum \frac{1}{n^3}$ and $\sum n$. Then we have that $\sum \frac{1}{n^3} \cdot n = \sum \frac{1}{n^2}$, which is a convergent $p$ series with $p=2>1$. However, we have that while $\sum \frac{1}{n^3}$ converges since it is a convergent $p$-series with $p=3>1$, but $\sum n$ diverges.\\
+			
+			\item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_n$ and $\sum b_n$ converge, then $\sum\sqrt{(a_n)^2+(b_n)^2}$ converges.
+			\\\\This is a true statement.
+			\begin{proof}
+				Let $\sum a_n$ and $\sum b_n$ be convergent series, and let $a_n$ and $b_n$ be positive sequence of real numbers. Then by \textit{Theorem 3.7.3}, we know that the sequences of partial sums $(s_k)$ and $(t_k)$ of $\sum a_n$ and $\sum b_n$, respectively, must be bounded. And since the sequences of partial sums converges, we know that by the  \textit{Cauchy Convergence Criterion}, $(s_k)$ and $(t_k)$ are Cauchy sequences. Additionally, by \textit{Theorem 3.2.3}, since both $(s_k)$ and $(t_k)$ converge, then $(s_k)+(t_k)$ must also converge. This yields that $\forall\ \varepsilon>0,\ \exists\ K(\varepsilon) \in \N\ \st \forall\ n>m\geq K(\varepsilon),$ by the \textit{Triangle Inequality}, we have 
+				\[\sqrt{(s_n-s_m)^2+(t_n-t_m)^2} \leq \sqrt{(s_n-s_m)^2} + \sqrt{(t_n-t_m)^2} = |s_n-s_m|+|t_n-t_m| < \varepsilon\]
+				And thus since this is the definition of the \textit{Cauchy Criterion for Series}, we have that this is equal to the series $\sum \sqrt{(a_n)^2+(b_n)^2}$. Therefore the series $\sum \sqrt{(a_n)^2+(b_n)^2}$ converges.
+			\end{proof}
+		
+			\item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum \sqrt{(a_n)^2 + (b_n)^2}$ converges, then $\sum a_n$ and $\sum b_n$ converge.
+			\\\\This is a true statement.
+			\begin{proof}
+				By the \textit{Comparison Test}, since $0 \leq a_n^2 \leq a_n^2+b_n^2$, we have that $0 \leq a_n \leq \sqrt{a_n^2+b_n^2}$, and thus $\sum a_n$ converges. Similarly, by the \textit{Comparison Test}, we have that since $0 \leq b_n^2 \leq a_n^2+b_n^2$, we have that $0 \leq b_n \leq \sqrt{a_n^2+b_n^2}$, and thus $\sum b_n$ converges.
+			\end{proof}
+			
+			\item If $\sum a_n$ converges and $0 \leq b_n \leq a_n$ , then $\sum b_n$ converges.
+			\\\\This is true since it is the \textit{Comparison Test}.\\
+			
+			\item If $\limx{n}{\infty} a_n=0,\ a_n \geq 0$ and $\sqrt{a_{n+1}} \leq a_n$ for all $n \in \N$, then $\sum a_n$ converges.
+			\begin{proof}
+				Since $\lim a_n = 0$, we know that $\exists\ N \in \N \st \forall\ n \geq N,\ |a_n|\leq \frac{1}{2}$. (That is, let $\varepsilon = \frac{1}{2}$). Also, since $\sqrt{a_{n+1}} \leq a_n$, we know that $a_{n+1} \leq a_N^2 \leq \frac{1}{4}$. So $a_{n+1} \leq \frac{1}{4}$. So, we have
+				\begin{align*}
+					a_{n+2} &\leq a_{n+1}^2\leq \frac{1}{16} = \frac{1}{4^2} \\
+					&\dots \\
+					a_{n+3} &\leq \frac{1}{4^4}
+				\end{align*}
+				So, we have that
+				\[|a_n| \leq \left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots+\frac{1}{4^{n-N}}\right)\]
+				since $a_n=a_{N(n-N)}$. So, we have that $0 \leq |a_n| \leq \frac{1}{4}$, which is a convergent geometric series. Therefore by the \textit{Comparison Test}, we have that $\sum a_n$ is convergent.
+			\end{proof}
+		\end{enumerate}
+	\end{enumerate}
+\end{document}

Documents/LaTeX/Homework 8.tex

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+\documentclass[12pt,letterpaper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{pgfplots}
+\usepackage[english]{babel}
+\usepackage{amsthm}
+\usepackage{cancel}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage{array}
+\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
+\usepackage{enumitem}
+\usepackage{mathrsfs}
+\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}}
+\newcommand{\st}{\ \text{s.t.}\ }
+\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
+\newcommand{\R}{\mathbb{R}}
+\newcommand{\N}{\mathbb{N}}
+\newcommand{\Q}{\mathbb{Q}}
+\newcommand{\C}{\mathbb{C}}
+\newcommand{\Z}{\mathbb{Z}}
+\newcommand{\dotp}{\dot{\mathcal{P}}}
+\newcommand{\dotq}{\dot{\mathcal{Q}}}
+\newcommand{\dist}{\text{dist}}
+\DeclareMathOperator{\sign}{sgn}
+\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
+\theoremstyle{case}
+\newtheorem{case}{Case}
+\newtheorem{case*}{Case}
+\theoremstyle{definition}
+\newtheorem{definition}{Definition}[section]
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem*{theorem*}{Theorem}
+\newtheorem{corollary}{Corollary}[section]
+\newtheorem*{corollary*}{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem*{lemma*}{Lemma}
+\newtheorem*{remark}{Remark}
+\setlist[enumerate]{font=\bfseries}
+\renewcommand{\qedsymbol}{$\blacksquare$}
+\author{Alexander J. Tusa}
+\title{Real Analysis II Homework 8}
+\begin{document}
+	\maketitle
+	\begin{enumerate}
+		\item \textbf{Section 9.2}
+		\begin{enumerate}
+			\item[2. (c)] Establish the convergence or divergence of the series whose $n$th term is $n!/n^n$.
+			\\\\Since
+			\begin{align*}
+				\abs{\frac{a_{n+1}}{a_n}} &= \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \\
+				&= \frac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!} \\
+				&= \frac{n^n}{(n+1)^n} \\
+				&\Downarrow \\
+				\limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}} &= \frac{1}{\limx{n}{\infty} \left(1+\frac{1}{n}\right)^n} \\
+				&= \frac{1}{e} \\
+				&<1
+			\end{align*}
+			By the \textit{Ratio Test}, and by \textit{Corollary 9.2.5}, we have that $\sum \frac{n!}{n^n}$ converges.\\
+			
+			\item[5.] Show that the series $1/1^2+1/2^3+1/3^2+1/4^3+\dots$ is convergent, but that both the Ratio and the Root Tests fail to apply.
+			\\\\We notice that the specified series yields $S=\sum a_n$, where
+			\[a_{2n}=\frac{1}{(2n)^3},\ \text{ and }\ a_{2n-1}=\frac{1}{(2n-1)^2},\ \text{ for }\ n \in \N\]
+			\\\\First we show that the \textit{Ratio Test} fails. To do so, we must consider two cases:
+			\begin{align*}
+				\abs{\frac{a_{2n+1}}{a_{2n}}} &= \frac{\frac{1}{(2n+1)^2}}{\frac{1}{(2n)^3}} \\
+				&= \frac{8n^3}{4n^2+4n+1} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\
+				&= \frac{8}{\frac{4}{n}+\frac{4}{n^2}+\frac{1}{n^3}} \\
+				&\Downarrow \\
+				\limx{n}{\infty} \frac{8}{\frac{4}{n}+\frac{4}{n^2}+\frac{1}{n^3}} = \infty
+			\end{align*}
+			and
+			\begin{align*}
+				\abs{\frac{a_{2n}}{a_{2n-1}}} &= \frac{\frac{1}{(2n)^3}}{\frac{1}{(2n-1)^2}} \\
+				&= \frac{4n^2-4n+1}{8n^3} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\
+				&= \frac{\frac{4}{n}-\frac{4}{n^2}+\frac{1}{n^3}}{8} \\
+				&\Downarrow \\
+				\limx{n}{\infty} \frac{\frac{4}{n}-\frac{4}{n^2}+\frac{1}{n^3}}{8} &= 0
+			\end{align*}
+			And thus we can see that the \textit{Ratio Test} is ineffective on this series.
+			\\\\As for the \textit{Root Test}, we must also consider two cases:
+			\[\abs{a_{2n}}^{\frac{1}{2n}}=\left(\frac{1}{8n^3}\right)^{\frac{1}{2n}}=\frac{1}{8^\frac{2}{n}}\cdot \left(\frac{1}{n^{\frac{2}{n}}}\right)^3\implies \limx{n}{\infty} \frac{1}{8^\frac{2}{n}}\cdot \left(\frac{1}{n^{\frac{2}{n}}}\right)^3 = \frac{1}{1} \cdot \left(\frac{1}{1}\right)^3=1\]
+			and
+			\[\abs{a_{2n-1}}^{\frac{1}{2n-1}}=\left(\frac{1}{(2n-1)^2}\right)^{\frac{1}{2n-1}}=\left(\frac{1}{2n-1}\right)^{\frac{2}{2n-1}} \implies \limx{n}{\infty} \left(\frac{1}{2n-1}\right)^{\frac{2}{2n-1}}=1\]
+			And thus we see that the \textit{Root Test} is ineffective on this series as well.
+			\\\\Now, we'll show that the series does converge by the \textit{Comparison Test}:
+			\\\\We notice that
+			\[a_{2n}=\frac{1}{8n^3}<\frac{1}{n^3}<\frac{1}{n^2}\]
+			\[a_{2n-1}=\frac{1}{(2n-1)^2}<\frac{1}{n^2}\]
+			and thus by the \textit{Comparison Test}, we have that since $\frac{1}{n^2}$ is a convergent $p$-series with $p=2>1$, the series $\sum a_n$ must also converge.\\
+			
+			\item[7.] Discuss the series whose $n$th term is
+			\begin{enumerate}
+				\item[(a)] $\displaystyle\frac{n!}{3\cdot5\cdot7\cdot\dots\cdot(2n+1)}$
+				\\\\By the \textit{Ratio Test}, we have:
+				\begin{align*}
+					\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{(n+1)!}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2(n+1)+1)}}{\displaystyle\frac{n!}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}} \\
+					&= \frac{(n+1)!\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}{n! \cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)\cdot(2n+3)} \\
+					&= \frac{n+1}{2n+3} \leq \frac{n+1}{2n+2} \\
+					&= \frac{n+1}{2(n+1)} \\
+					&= \frac{1}{2} \\
+					&< 1
+				\end{align*}
+				And thus by the \textit{Ratio Test}, we have that the series is absolutely convergent.\\
+				
+				\item[(b)] $\displaystyle\frac{(n!)^2}{(2n)!}$
+				\\\\By the \textit{Ratio test}, we have:
+				\begin{align*}
+					\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{\big((n+1)!\big)^2}{\big(2(n+1)\big)!}}{\displaystyle\frac{(n!)^2}{(2n)!}} \\
+					&= \frac{(n+1)! \cdot (n+1)! \cdot (2n)!}{(2n+2)!\cdot n! \cdot n!} \\
+					&= \frac{(n+1)(n+1)}{(2n+1)(2n+2)} \\
+					&= \frac{n+1}{2(2n+1)} \\
+					&= \frac{n+1}{4n+2} \\
+					&\leq \frac{n+1}{4n} \\
+					&= \frac{1}{4} + \frac{1}{4n} \\
+					&\leq \frac{1}{4}+\frac{1}{4} \\
+					&= \frac{1}{2} \\
+					&< 1
+				\end{align*}
+				And thus by the \textit{Ratio Test}, we have that the series is absolutely convergent.\\
+				
+				\item[(c)] $\displaystyle\frac{2\cdot4\cdot\dots\cdot(2n)}{3\cdot5\cdot\dots\cdot(2n+1)}$
+				\\\\By the \textit{Ratio Test}, we have:
+				\begin{align*}
+					\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n) \cdot (2n+2)}{3 \cdot 5 \cdot \dots \cdot (2n+1) \cdot (2n+3)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{3 \cdot 5 \cdot \dots \cdot (2n+1)}} \\
+					&= \frac{2n+2}{2n+3} \\
+					&\Downarrow \\
+					\limx{n}{\infty} \abs{\frac{2n+2}{2n+3}} &= 1
+				\end{align*}
+				Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series.
+				\\\\By \textit{Raabe's Test}, we have:
+				\begin{align*}
+					\abs{\frac{a_{n+1}}{a_n}} &= \frac{2n+2}{2n+3} \\
+					&= \frac{(2n+3)-1}{2n+3} \\
+					&= 1-\frac{1}{2n+3} \\
+					&\geq 1-\frac{1}{2n} \\
+					&= 1-\frac{\frac{1}{2}}{n}
+				\end{align*}
+				Thus by \textit{Raabe's Test}, since $a=\frac{1}{2}$, we have that the series is divergent.\\
+				
+				\item[(d)] $\displaystyle\frac{2\cdot4\cdot\dots\cdot(2n)}{5\cdot7\cdot\dots\cdot(2n+3)}$
+				\\\\By the \textit{Ratio Test}, we have:
+				\begin{align*}
+					\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)\cdot (2n+2)}{5 \cdot 7 \cdot \dots \cdot (2n+3) \cdot (2n+5)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{5 \cdot 7 \cdot \dots \cdot (2n+3)}} \\
+					&= \frac{2n+2}{2n+5} \\
+					&\Downarrow \\
+					\limx{n}{\infty} \frac{2n+2}{2n+5} &= 1
+				\end{align*}
+				Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series.
+				\\\\By \textit{Corollary 9.2.9}, we have:
+				\begin{align*}
+					a &= \limx{n}{\infty} \left(n \left(1-\abs{\frac{a_{n+1}}{a_n}}\right)\right) \\
+					&= \limx{n}{\infty} \left(n \left(1-\frac{2n+2}{2n+5}\right)\right) \\
+					&= \limx{n}{\infty} \left(n \cdot \frac{3}{2n+5}\right) \\
+					&= \limx{n}{\infty} \left(\frac{3}{2+\frac{5}{n}}\right) \\
+					&= \frac{3}{2}
+				\end{align*}
+				Thus by \textit{Corollary 9.2.9}, since $a=\frac{3}{2} > 1$, we have that the series is absolutely convergent.
+			\end{enumerate}
+		\end{enumerate}
+	
+		\item \textbf{Section 9.3}
+		\begin{enumerate}
+			\item[1.] Test the following series for convergence and for absolute convergence:
+			\begin{enumerate}
+				\item[(a)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2+1}$
+				\\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n^2+1}=0$. And by the \textit{Limit Comparison Test}, we notice that $\frac{1}{n^2+1}$ looks like $\frac{1}{n^2}$, which we note is a convergent $p$-series with $p=2>1$, which yields:
+				\[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n^2+1}}{\displaystyle\frac{1}{n^2}}= \limx{n}{\infty} \frac{n^2}{n^2+1} = 1 \neq 0\]
+				And thus since $\sum \frac{1}{n^2}$ is convergent, by the \textit{Limit Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2+1}$ is absolutely convergent.\\
+				
+				\item[(b)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$
+				\\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n+1} = 0$, and thus the series is convergent. And by the \textit{Limit Comparison Test}, we note that the series looks like $\sum \frac{1}{n}$, which we note is a harmonic series and thus diverges, which yields
+				\[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n+1}}{\frac{1}{n}} = \limx{n}{\infty} \frac{n}{n+1} = 1 \neq 0\]
+				And since $\sum \frac{1}{n}$ is a harmonic series and thus diverges, since the limit is not equal to 0, we have that by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is conditionally convergent.\\
+				
+				\item[(c)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n+2}$
+				\\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{n}{n+2} = 1 \neq 0$, and thus the series is divergent.\\
+				
+				\item[(d)] $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\ln n}{n}$
+				\\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{\ln n}{n} = 0$, which yields that the series is convergent. And by the \textit{Integral Test}, we have that
+				\[\int_{1}^{\infty} \frac{\ln n}{n}\ dn = \int_{1}^{\infty} \frac{u}{n}\cdot n\ du = \int_{1}^{\infty} u\ du = \left.\frac{u^2}{2}\right|_1^\infty = \left.\frac{(\ln n)^2}{2}\right|_1^\infty = \frac{(\ln \infty)^2}{2} - \frac{(\ln 1)^2}{2}\]
+				\[= \infty - 0 = \infty\]
+				Thus by the \textit{Integral Test}, we have that this series is divergent. Thus the series is conditionally convergent.\\
+				
+			\end{enumerate}
+		
+			\item[3.] Give an example to show that the Alternating Series Test 9.3.2 may fail if $(z_n)$ is not a decreasing sequence.
+			\\\\Let $\sum a_n$ be the series defined as
+			\[\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\dots+\frac{1}{n}-\frac{1}{2n}+\dots\right)\]
+			Let $A_n$ be the partial sums of the series $\sum a_n$. Then since $a_n$ is an alternating sequence that converges to 0 and isn't decreasing, we have
+			\[A_{2n}=\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{2n}\right) = \sum_{n=1}^{\infty} \left(\frac{1}{2n}\right)\]
+			As $A_{2n}$ diverges, we have that $A_n$ diverges and thus the series $\sum a_n$ is divergent.\\
+			
+			\item[5.] Consider the series
+			\[1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}++--\dots,\]
+			where the signs come in pairs. Does it converge?
+			\\\\We notice that $\frac{1}{n}$ is a monotone decreasing sequence that converges to 0. Then we notice that the series $\displaystyle\sum_{n=1}^{\infty} a_n$ where for every $n \in \N$, we have $a_1=1, a_{4n}=1, a_{4n-1}=-1=a_{4n+1}$. Now, let $s_n=a_1+a_2+\dots+a_n$. Then $s_{2n}=0$ and $s_{2n+1}=\pm 1$. This yields that $|s_n| \leq 1$. By \textit{Dirichlet's Test}, we have that $\displaystyle\sum_{n=1}^{\infty} a_n$ is convergent. Thus we have
+			\[\sum_{n=1}^{\infty} \frac{a_n}{n}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}++--\dots\]
+			is also convergent.
+		\end{enumerate}
+	
+		\item Give an example of a series $\sum a_n$ that consists of nonzero terms with $\limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}}=1$ for each of the following conditions:
+		\begin{enumerate}
+			\item $\sum a_n$ converges absolutely
+			\\\\Consider Problem 7d from Section 9.2: $\displaystyle\sum_{n=1}^{\infty}\frac{2\cdot4\cdot\dots\cdot(2n)}{5\cdot7\cdot\dots\cdot(2n+3)}$
+			\\\\By the \textit{Ratio Test}, we have:
+			\begin{align*}
+			\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)\cdot (2n+2)}{5 \cdot 7 \cdot \dots \cdot (2n+3) \cdot (2n+5)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{5 \cdot 7 \cdot \dots \cdot (2n+3)}} \\
+			&= \frac{2n+2}{2n+5} \\
+			&\Downarrow \\
+			\limx{n}{\infty} \frac{2n+2}{2n+5} &= 1
+			\end{align*}
+			Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series.
+			\\\\By \textit{Corollary 9.2.9}, we have:
+			\begin{align*}
+			a &= \limx{n}{\infty} \left(n \left(1-\abs{\frac{a_{n+1}}{a_n}}\right)\right) \\
+			&= \limx{n}{\infty} \left(n \left(1-\frac{2n+2}{2n+5}\right)\right) \\
+			&= \limx{n}{\infty} \left(n \cdot \frac{3}{2n+5}\right) \\
+			&= \limx{n}{\infty} \left(\frac{3}{2+\frac{5}{n}}\right) \\
+			&= \frac{3}{2}
+			\end{align*}
+			Thus by \textit{Corollary 9.2.9}, since $a=\frac{3}{2} > 1$, we have that the series is absolutely convergent.\\
+			
+			\item $\sum a_n$ converges conditionally
+			\\\\Consider problem 1b from Section 9.3: $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$
+			\\\\By the \textit{Ratio Test}, we have:
+			\begin{align*}
+			\limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}} &= \limx{n}{\infty} \frac{\displaystyle\frac{1}{n+3}}{\displaystyle\frac{1}{n+1}} \\
+			&= \limx{n}{\infty} \frac{n+1}{n+3} \\
+			&= 1
+			\end{align*}
+			Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series.
+			\\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n+1} = 0$, and thus the series is convergent. And by the \textit{Limit Comparison Test}, we note that the series looks like $\sum \frac{1}{n}$, which we note is a harmonic series and thus diverges, which yields
+			\[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n+1}}{\frac{1}{n}} = \limx{n}{\infty} \frac{n}{n+1} = 1 \neq 0\]
+			And since $\sum \frac{1}{n}$ is a harmonic series and thus diverges, since the limit is not equal to 0, we have that by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is conditionally convergent.\\
+			
+			\item $\sum a_n$ diverges.
+			\\\\Consider Problem 7c from Section 9.2: $\displaystyle\sum_{n=1}^{\infty}\frac{2\cdot4\cdot\dots\cdot(2n)}{3\cdot5\cdot\dots\cdot(2n+1)}$
+			\\\\By the \textit{Ratio Test}, we have:
+			\begin{align*}
+			\abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n) \cdot (2n+2)}{3 \cdot 5 \cdot \dots \cdot (2n+1) \cdot (2n+3)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{3 \cdot 5 \cdot \dots \cdot (2n+1)}} \\
+			&= \frac{2n+2}{2n+3} \\
+			&\Downarrow \\
+			\limx{n}{\infty} \abs{\frac{2n+2}{2n+3}} &= 1
+			\end{align*}
+			Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series.
+			\\\\By \textit{Raabe's Test}, we have:
+			\begin{align*}
+			\abs{\frac{a_{n+1}}{a_n}} &= \frac{2n+2}{2n+3} \\
+			&= \frac{(2n+3)-1}{2n+3} \\
+			&= 1-\frac{1}{2n+3} \\
+			&\geq 1-\frac{1}{2n} \\
+			&= 1-\frac{\frac{1}{2}}{n}
+			\end{align*}
+			Thus by \textit{Raabe's Test}, since $a=\frac{1}{2}$, we have that the series is divergent.\\
+			
+		\end{enumerate}
+	
+		\item Prove or justify, if true. Provide a counterexample, if false.
+		\begin{enumerate}
+			\item If $\sum |a_n|$ diverges, then $\sum a_n$ is conditionally convergent.
+			\\\\This is a false statement. Consider the sequence $a_n=(0,\frac{1}{2},0,\frac{1}{4},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows:
+			\[a_n:=\begin{cases}
+			0, &n\text{ is odd} \\
+			\frac{1}{n}, &n\text{ is even}
+			\end{cases}\]
+			for all $n \in \N$. By this definition, we have
+			\begin{align*}
+				\sum_{n=1}^{\infty} |a_n| &= \sum_{n=1}^{\infty} a_n \\
+				&= 0+\frac{1}{2}+0+\frac{1}{4}+0+\frac{1}{6}+\dots \\
+				&= \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots \\
+				&= \sum_{n=1}^{\infty} \frac{1}{2n} \\
+				&= \infty
+			\end{align*}
+			Thus we have that $\sum |a_n|$ diverges since it yields a harmonic series, and since all terms of $a_n$ are greater than or equal to 0, we have that the series $\sum |a_n|=\sum a_n$. Thus $\sum a_n$ also diverges.\\
+			
+			\item If $\sum |a_n|$ diverges, then $\sum |a_n|$ is conditionally convergent.
+			\\\\This is a false statement. Refer to the previous problem's counterexample: Consider the sequence $a_n=(0,\frac{1}{2},0,\frac{1}{4},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows:
+			\[a_n:=\begin{cases}
+			0, &n\text{ is odd} \\
+			\frac{1}{n}, &n\text{ is even}
+			\end{cases}\]
+			for all $n \in \N$. By this definition, we have
+			\begin{align*}
+			\sum_{n=1}^{\infty} |a_n| &= \sum_{n=1}^{\infty} a_n \\
+			&= 0+\frac{1}{2}+0+\frac{1}{4}+0+\frac{1}{6}+\dots \\
+			&= \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots \\
+			&= \sum_{n=1}^{\infty} \frac{1}{2n} \\
+			&= \infty
+			\end{align*}
+			Thus we have that $\sum |a_n|$ diverges since it yields a harmonic series, and since all terms of $a_n$ are greater than or equal to 0, we have that the series $\sum |a_n|=\sum a_n$. Thus $\sum a_n$ also diverges.\\
+			
+			\item If $\sum |a_n|$ diverges, then $\sum a_n$ diverges.
+			\\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. Then, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2$, but $\displaystyle\sum_{n=1}^{\infty} \abs{\frac{(-1)^{n+1}}{n}} = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = \infty$, since this is a harmonic series. Thus we have that $\sum |a_n|$ diverges since it yields the harmonic series, but $\sum a_n$ converges to $\ln 2$, hence $\sum a_n$ is conditionally convergent.\\
+			
+			\item If $\sum |a_n|$ converges, then $\sum a_n$ is absolutely convergent.
+			\\\\This is true since it is the definition of \textit{Absolute Convergence}.\\
+			
+			\item If $a_n \leq b_n$ for all $n \in \N$ and $\sum b_n$ is absolutely convergent, then $\sum a_n$ converges.
+			\\\\This is true by the \textit{Comparison Test}.\\
+			
+			\item If $\sum a_n$ is absolutely convergent, then $\sum a_n^2$ is absolutely convergent.
+			\\\\This is a true statement.
+			\begin{proof}
+				Let $\sum a_n$ be an absolutely convergent series. Then $\limx{n}{\infty} a_n=0$ by the \textit{$n$th Term Test}. Thus, we know that $\exists\ N \in \N \st 0 < a_n < 1,\ \forall\ n \geq N$, and thus $0 < a_n^2<a_n<1,\ \forall\ n \geq N$, and thus by the \textit{Comparison Test}, the series $\sum a_n^2$ is also absolutely convergent.
+			\end{proof}
+			
+			\item If $\lim a_n = 0$, then $\sum (-1)^na_n$ converges.
+			\\\\This is a false statement. Consider the example given in Problem 3 of Section 9.3:
+			\\\\Let $\sum a_n$ be the series defined as
+			\[\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\dots+\frac{1}{n}-\frac{1}{2n}+\dots\right)\]
+			Let $A_n$ be the partial sums of the series $\sum a_n$. Then since $a_n$ is an alternating sequence that converges to 0 and isn't decreasing, we have
+			\[A_{2n}=\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{2n}\right) = \sum_{n=1}^{\infty} \left(\frac{1}{2n}\right)\]
+			As $A_{2n}$ diverges, we have that $A_n$ diverges and thus the series $\sum a_n$ is divergent.\\
+			
+			\item If $\lim a_n=0$ and $a_n \geq 0$ for all $n \in \N$, then $\sum (-1)^n a_n$ converges.
+			\\\\This is a false statement. Consider the sequence $a_n=(1,0,\frac{1}{3},0,\frac{1}{5},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows:
+			\[a_n:=\begin{cases}
+			0, &n\text{ is even} \\
+			\frac{1}{n}, &n\text{ is odd}
+			\end{cases}\]
+			for all $n \in \N$. By this definition, we have that $\lim a_n=0$.
+			\\\\However, notice 
+			\begin{align*}
+				\sum_{n=1}^{\infty}(-1)^n a_n &= -1 + 0 -\frac{1}{3}+0-\frac{1}{5}+0-\frac{1}{7}+0-+-+\dots \\
+				&= -1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\dots \\
+				&= -\sum_{n=1}^{\infty} \frac{1}{2n-1} \\
+				&= -\infty
+			\end{align*}
+			Yields a negative harmonic series, which diverges. Hence $\sum (-1)^na_n$ diverges. \\
+			
+			\item If $\lim a_n=0$ and $\sum (-1)^n a_n$ converges, then $a_n$ is decreasing.
+			\\\\This is a false statement. Consider the sequence $a_n=(0,\frac{1}{4},0,\frac{1}{16},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows:
+			\[a_n:=\begin{cases}
+			0, &n\text{ is odd} \\
+			\frac{1}{n^2}, &n\text{ is even}
+			\end{cases}\]
+			for all $n \in \N$. By this definition, we have that $\lim a_n=0$.
+			\\\\However, notice 
+			\begin{align*}
+			\sum_{n=1}^{\infty}(-1)^n a_n &= -0 +\frac{1}{4}-0+\frac{1}{16}-0+\frac{1}{36}-0+-+-\dots \\
+			&= \frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\dots \\
+			&= \frac{1}{4} \left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\dots\right) \\
+			&= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\
+			&= \frac{1}{4} \cdot \frac{\pi^2}{6} \\
+			&= \frac{\pi^2}{24}
+			\end{align*}
+			Thus we have that $\sum (-1)^na_n$ converges. However, the sequence $a_n$ is not a decreasing sequence. Thus $\lim a_n=0$, and $\sum (-1)^na_n$ converges, but $a_n$ is not decreasing.
+			
+			\item If $a_n \geq 0$ for all $n$ and $\sum a_n$ converges, then $\sum \sin a_n$ converges.
+			\\\\This is a false statement. Consider $a_n=(0,0,0,0,\dots)$. Then we have that $\sum a_n=0+0+0+0+\dots=0$. Thus $\sum a_n$ converges. However, $\sum \sin a_n= \sin(0)+\sin(0)+\sin(0)+\sin(0)+\dots = 1 + 1 + 1 + 1 + \dots = \infty$, and thus $\sum \sin a_n$ diverges.\\
+		\end{enumerate}
+	
+		\item Assume that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$. Prove that:
+		\begin{enumerate}
+			\item $\displaystyle \sum \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$.
+			\begin{align*}
+				\sum \frac{1}{n^2} &= 1+\frac{1}{2^2} +\frac{1}{3^2}+\frac{1}{4^2} + \frac{1}{5^2}+\frac{1}{6^2} \dots \\
+				&= 1+\frac{1}{3^2}+\frac{1}{5^2}+\dots+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\dots \\
+				&= \sum \frac{1}{(2n-1)^2} + \frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots\right) \\
+				&= \sum \frac{1}{(2n-1)^2} + \frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots\right) \\
+				&= \sum \frac{1}{(2n-1)^2} + \frac{1}{4} \sum \frac{1}{n^2} \\
+				&= \sum \frac{1}{(2n-1)^2} + \frac{1}{4} \cdot \frac{\pi^2}{6} \\
+				&= \sum \frac{1}{(2n-1)^2} + \frac{\pi^2}{24} \\
+				&\Downarrow \\
+				\sum \frac{1}{(2n-1)^2} &= \sum \frac{1}{n^2} - \frac{1}{4} \sum \frac{1}{n^2} \\
+				&= \frac{\pi^2}{6} - \frac{\pi^2}{24} \\
+				&= \frac{\pi^2}{8}
+			\end{align*}
+			
+			\item $\displaystyle\frac{\pi^2}{24}=\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}+\dots$
+			\begin{align*}
+				\sum \frac{1}{(2n)^2} &= \frac{1}{2^2} + \frac{1}{4^2}+\frac{1}{6^2}+ \dots \\
+				&= \frac{1}{4} \left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots\right) \\
+				&= \frac{1}{4} \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots\right) \\
+				&= \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} \\
+				&= \frac{1}{4} \cdot \frac{\pi^2}{6} \\
+				&= \frac{\pi^2}{24}
+			\end{align*}
+			
+			\item $\displaystyle \frac{\pi^2}{12} = 1 -\frac{1}{2^2}+\frac{1}{3^2} - \frac{1}{4^2} + \dots$
+			\begin{align*}
+				\sum \frac{(-1)^{n+1}}{n^2} &= 1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}-\frac{1}{6^2}+\dots \\
+				&= 1+\frac{1}{3^2}+\frac{1}{5^2}+\dots+\frac{1}{(2n-1)^2} -\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{6^2}-\dots-\frac{1}{(2n)^2} \\
+				&= \sum \frac{1}{(2n-1)^2} -\left(\frac{1}{2^1}+\frac{1}{4^2}+\frac{1}{6^2}+\dots+\frac{1}{(2n)^2}\right) \\
+				&= \sum \frac{1}{(2n-1)^2} - \sum \frac{1}{(2n)^2} \\
+				&= \frac{\pi^2}{8} -\frac{\pi^2}{24} \\
+				&= \frac{\pi^2}{12}
+			\end{align*}
+			By our previous answers for part (a) and (b).
+		\end{enumerate}
+	\end{enumerate}
+\end{document}