\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