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chapter-3/chapter-3.tex
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\chapter{Sequences and Series}
\subimport{./}{sequences-and-their-limits.tex}
\subimport{./}{limit-theorems.tex}
\subimport{./}{monotone-sequences.tex}
\subimport{./}{subsequences-and-the-bolzano-wierstrass-theorem.tex}
\subimport{./}{the-cauchy-criterion.tex}
\subimport{./}{properly-divergent-sequences.tex}
\subimport{./}{introduction-to-infinite-series.tex}
chapter-3/introduction-to-infinite-series.tex
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\section{Introduction to Infinite Series}
\begin{definition}
If $X:=(x_n)$ is a sequence in $\R$, then the \textbf{infinite series} (or simply the \textbf{series}) \textbf{generated by} $X$ is the sequence $S:= (s_k)$ defined by
\begin{align*}
s_1 & := x_1 \\
s_2 & := s_1 + s_2 & (=x_1 + x_2) \\
& \dots \\
s_k & := s_{k-1}+x_k & (=x_1+x_2+\dots+x_k) \\
& \dots
\end{align*}
\end{definition}
\begin{theorem}[\textbf{The $n$th Term Test}]
If the series $\sum x_n$ converges, then $\lim (x_n) = 0$.
\end{theorem}
\begin{theorem}[\textbf{Cauchy Criterion for Series}]
The series $\sum x_n$ converges if and only if for every $\varepsilon >0$ there exists $M(\varepsilon) \in \N$ such that if $m>n\geq M(\varepsilon)$, then
\[|s_m-s_n|=|x_{n+1}+x_{n+2}+\dots+x_m|<\varepsilon\]
\end{theorem}
\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,
\[\sum\limits_{n=1}^{\infty}x_n = \lim (x_k) = \sup \{s_k:k \in \N\}\]
\end{theorem}
\begin{theorem}[\textbf{Comparison Test}]
Let $X:=(x_n)$ and $Y:=(y_n)$ be real sequences and suppose that for some $K \in \N$ we have
\[0 \leq x_n \leq y_n\ \ \ \text{for}\ \ \ n \geq K\]
\begin{enumerate}
\item Then the convergence of $\sum y_n$ implies the convergence of $\sum x_n$.
\item The divergence of $\sum x_n$ implies the divergence of $\sum y_n$.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Limit Comparison Test}]
Suppose that $X:=(x_n)$ and $Y:=(y_n)$ are strictly positive sequences and suppose that the following limit exists in $\R$:
\[r:=\lim \left(\frac{x_n}{y_n}\right)\]
\begin{enumerate}
\item If $r \neq 0$, then $\sum x_n$ is convergent if and only if $\sum y_n$ is convergent.
\item If $r=0$ and if $\sum y_n$ is convergent, then $\sum x_n$ is convergent.
\item If $r=\infty$ and $\sum y_n$ diverges, then $\sum x_n$ diverges.
\end{enumerate}
\end{theorem}
\begin{definition}
Let $(a_n):n \mapsto a(n)$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero. That is, for every $n$ in the domain of $(a_n):a_n>0,\ a_{n+1} \leq a_n$, and $a_n \to 0$ as $n \to + \infty$. The series $\displaystyle\sum_{n=1}^{\infty} 2^n a(2^n)$ is called the \textbf{condensed} form of the series $\displaystyle\sum_{n=1}^{\infty} a_n$.
\end{definition}
\begin{theorem}[\textbf{Cauchy Condensation Test}]
Let $(a_n):n \mapsto a(n)$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero. That is, for every $n$ in the domain of $(a_n):a_n>0,\ a_n \geq a_{n+1}$, and $a_n \to 0$ as $n \to +\infty$. Then the series $\displaystyle\sum_{n=1}^{\infty} a_n$ converges if and only if the condensed series $\displaystyle\sum_{n=1}^{\infty} 2^na(2^n)$ converges.
\end{theorem}
\begin{theorem}[\textbf{Cauchy Ratio Test}]
Let $\displaystyle\sum_{n=1}^{\infty} a_n$ be a series and $a_n>0$ for all $n \in \N$, and suppose the following limit exists in $\R$:
\[L=\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}\]
\begin{enumerate}
\item If $L<1$, then the series converges absolutely.
\item If $L>1$, then the series is divergent.
\item If $L=1$ or the limit does not exist, then the test is inconclusive.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Kummer's Test}]
Let $\sum a_n$ be a positive term series.
\begin{enumerate}
\item If there exists a positive term sequence $b_n$, $\alpha > 0$, and $\N \in \N$ such that $\displaystyle\frac{a_n}{a_{n+1}} \cdot b_n - b_{n+1} \geq \alpha,\ \forall\ n \geq N$, then $\sum a_n$ converges.
\item If $\displaystyle\frac{a_n}{a_{n+1}}\cdot b_n - b_{n+1} \leq 0,\ \forall\ n \geq N$, and if $\sum \frac{1}{b_n}$ diverges, then $\sum a_n$ diverges.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Gauss' Test}]
If $\sum a_n$ is a positive term series, and if there exists a bounded sequence $b_n$ such that $\forall\ n \geq N$, $\displaystyle\frac{a_n}{a_{n+1}} = 1 +\frac{L}{n} + \frac{b_n}{n^2}$, then
\begin{enumerate}
\item If $L > 1$, then $\sum a_n$ converges.
\item If $L \leq 1$, then $\sum a_n$ diverges.
\end{enumerate}
\end{theorem}
chapter-3/limit-theorems.tex
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\section{Limit Theorems}
\begin{definition}
A sequence $X=(x_n)$ of real numbers is said to be \textbf{bounded} if there exists a real number $M>0$ such that $|x_n|\leq M$ for all $n \in \N$.
\\\\Thus, the sequence $(x_n)$ is bounded if and only if the set $\{x_n : n \in \N\}$ of its values is a bounded subset of $\R$.
\end{definition}
\begin{theorem}
A convergent sequence of real numbers is bounded.
\end{theorem}
\begin{theorem}
Let $X=(x_n)$ and $Y=(y_n)$ be sequences of real numbers that converge to $x$ and $y$, respectively, and let $c \in \R$.
\begin{enumerate}
\item Then the sequences $X+Y, X-Y, X \cdot Y$, and $cX$ converge to $x+y, x-y, xy$, and $cx$, respectively.
\item If $X=(x_n)$ converges to $x$ and $Z=(z_n)$ is a sequence of nonzero real numbers that converges to $z$ and if $z \neq 0$, then the quotient sequence $X/Z$ converges to $x/z$.
\end{enumerate}
\end{theorem}
\begin{theorem}
If $X=(x_n)$ is a convergent sequence of real numbers and if $x_n \geq 0$ for all $n \in \N$, then $x = \lim (x_n) \geq 0$.
\end{theorem}
\begin{theorem}
If $X=(x_n)$ and $Y=(y_n)$ are convergent sequences of real numbers and if $x_n \leq y_n$ for all $n \in \N$, then $\lim (x_n) \leq \lim (y_n)$.
\end{theorem}
\begin{theorem}
If $X=(x_n)$ is a convergent sequence and if $a \leq x_n \leq b$ for all $n \in \N$, then $a \leq \lim (x_n) \leq b$.
\end{theorem}
\begin{theorem}[\textbf{Squeeze Theorem}]
Suppose that $X=(x_n), Y= (y_n)$, and $Z=(z_n)$ are sequences of real numbers such that
\[x_n \leq y_n \leq z_n\ \ \ \text{for all }\ \ \ n \in \N\]
and that $\lim (x_n) = \lim (z_n)$. Then $Y=(y_n)$ is convergent and
\[\lim (x_n) = \lim (y_n) = \lim (z_n).\]
\end{theorem}
\begin{theorem}
Let the sequence $X=(x_n)$ converge to $x$. Then the sequence $(|x_n|)$ of absolute values converges to $|x|$. That is, if $x=\lim (x_n)$, then $|x|=\lim (|x_n|)$.
\end{theorem}
\begin{theorem}
Let $X=(x_n)$ be a sequence of real numbers that converges to $x$ and suppose that $x_n \geq 0$. Then the sequence $(\sqrt{x_n})$ of positive square roots converges and $\lim (\sqrt{x_n})=\sqrt{x}$.
\end{theorem}
\begin{theorem}
Let $(x_n)$ be a sequence of positive real numbers such that $L := \lim (x_{n+1}/x_n)$ exists. If $L <1$, then $(x_n)$ converges and $\lim (x_n)=0$.
\end{theorem}
chapter-3/monotone-sequences.tex
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\section{Monotone Sequences}
\begin{definition}
Let $X=(x_n)$ be a sequence of real numbers. We say that $X$ is \textbf{increasing} if it satisfies the inequalities
\[x_1 \leq x_2 \leq \dots \leq x_n \leq x_{n+1} \leq \dots\]
We say that $X$ is \textbf{decreasing} if it satisfies the inequalities
\[x_1 \geq x_2 \geq \dots \geq x_n \geq x_{n+1} \geq \dots\]
We say that $X$ is \textbf{monotone} if it is either increasing or decreasing.
\end{definition}
\begin{theorem}[\textbf{Monotone Convergence Theorem}]
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}
chapter-3/properly-divergent-sequences.tex
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\section{Properly Divergent Sequences}
\begin{definition}
Let $(x_n)$ be a sequence of real numbers.
\begin{enumerate}
\item We say that $(x_n)$ \textbf{tends to} $\pm \infty$, and write $\lim (x_n) = +\infty$, if for every $\alpha \in \R$ there exists a natural number $K(\alpha)$ such that if $n \geq K(\alpha)$, then $x_n > \alpha$.
\item We say that $(x_n)$ \textbf{tends to} $-\infty$, and write $\lim (x_n) = -\infty$, if for every $\beta \in \R$ there exists a natural number $K(\beta)$ such that if $n \geq K(\beta)$, then $x_n < \beta$.
\end{enumerate}
We say that $(x_n)$ is \textbf{properly divergent} in case we have either $\lim (x_n)=+\infty$, or $\lim (x_n)=-\infty$.
\end{definition}
\begin{theorem}
A monotone sequence of real numbers is properly divergent if and only if it is unbounded.
\begin{enumerate}
\item If $(x_n)$ is an unbounded increasing sequence, then $\lim (x_n)=+\infty$.
\item If $(x_n)$ is an unbounded decreasing sequence, then $\lim (x_n) = -\infty$.
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $(x_n)$ and $(y_n)$ be two sequences of real numbers and suppose that
\[x_n \leq y_n\ \ \ \text{for all}\ \ \ n \in \N\]
\begin{enumerate}
\item If $\lim (x_n) = +\infty$, then $\lim (y_n)=+\infty$.
\item If $\lim (y_n) = -\infty$, then $\lim (x_n)=-\infty$.
\end{enumerate}
\end{theorem}
\begin{theorem}
Let $(x_n)$ and $(y_n)$ be two sequences of positive real numbers and suppose that for some $L \in \R, L>0$, we have
\[\lim (x_n/y_n)=L\]
Then $\lim (x_n)=+\infty$ if an only if $\lim (y_n)=+\infty$.
\end{theorem}
chapter-3/sequences-and-their-limits.tex
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\section{Sequences and Their Limits}
\begin{definition}
A \textbf{sequence of real numbers} (or a \textbf{sequence in $\R$}) is a function defined on the set $\N = \{1,2,\dots\}$ of natural numbers whose range is contained in the set $\R$ of real numbers.
\end{definition}
\begin{definition}
A sequence $X = (x_n)$ in $\R$ is said to \textbf{converge} to $x \in \R$, or $x$ is said to be a \textbf{limit} of $(x_n)$, if for every $\varepsilon >0$ there exists a natural number $K(\varepsilon)$ such that for all $n \geq K(\varepsilon)$, the terms $x_n$ satisfy $|x_n-x|<\varepsilon$.
\\If a sequence has a limit, we say that the sequence is \textbf{convergent}; if it has no limit, we say that the sequence is \textbf{divergent}.
\end{definition}
\begin{theorem}[\textbf{Uniqueness of Limits}]
A sequence in $\R$ can have at most one limit.
\end{theorem}
\begin{theorem}
Let $X = (x_n)$ be a sequence of real numbers, and let $x \in \R$. The following statements are equivalent:
\begin{enumerate}
\item $X$ converges $x$.
\item For every $\varepsilon > 0$, there exists a natural number $K$ such that for all $n \geq K$, the terms $x_n$ satisfy $|x_n -x| < \varepsilon$.
\item For every $\varepsilon >0$, there exists a natural number $K$ such that for all $n \geq K$, the terms $x_n$ satisfy $x-\varepsilon < x_n < x+\varepsilon$.
\item For every $\varepsilon$-neighborhood $V_\varepsilon(x)$ of $x$, there exists a natural number $K$ such that for all $n \geq K$, the terms $x_n$ belong to $V_\varepsilon(x)$.
\end{enumerate}
\end{theorem}
\begin{definition}
If $X=(x_1, x_2, \dots, x_n, \dots)$ is a sequence of real numbers and if $m$ is a given natural number, then the $m$-\textbf{tail} of $X$ is the sequence
\[X_m := (x_{m+n}: n \in \N)=(x_{m+1}, x_{m+2}, \dots)\]
\end{definition}
\begin{theorem}
Let $X=(x_n:n \in \N)$ be a sequence of real numbers and let $m \in \N$. Then the $m$-tail $X_m=(x_{m+n}:n \in \N)$ of $X$ converges if and only if $X$ converges. In this case, $\lim X_m = \lim X$.
\end{theorem}
\begin{theorem}
Let $(x_n)$ be a sequence of real numbers and let $x \in \R$. If $(a_n)$ is a sequence of positive real numbers with $\lim (a_n)=0$ and if for some constant $C >0$ and some $m \in \N$ we have
\[|x_n - x| \leq Ca_n\ \ \ \ \text{ for all }\ \ \ \ n \geq m\]
then it follows that $\lim (x_n) = x$.
\end{theorem}
chapter-3/subsequences-and-the-bolzano-wierstrass-theorem.tex
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\section{Subsequences and the Bolzano-Wierstrass Theorem}
\begin{definition}
Let $X=(x_n)$ be a sequence of real numbers and let $n_1 < n_2 < \dots < n_k < \dots$ be a strictly increasing sequence of natural numbers. Then the sequence $X' = (x_{n_k})$ given by
\[(x_{n_1}, x_{n_2}, \dots, x_{n_k}, \dots)\]
is called a \textbf{subsequence} of $X$.
\end{definition}
\begin{theorem}
If a sequence $X=(x_n)$ of real numbers converges to a real number $x$, then any subsequence $X' = (x_{n_k})$ of $X$ also converges to $x$.
\end{theorem}
\begin{theorem}
Let $X=(x_n)$ be a sequence of real numbers. Then the following are equivalent:
\begin{enumerate}
\item The sequence $X=(x_n)$ does not converge to $x \in \R$.
\item There exists an $\varepsilon_0 > 0$ such that for any $k \in \N$, there exists $n_k \in \N$ such that $n_k \geq k$ and $|x_{n_k}-x| \geq \varepsilon_0$.
\item There exists an $\varepsilon_0>0$ and a subsequence $X'=(x_{n_k})$ of $X$ such that $|x_{n_k}-x| \geq \varepsilon_0$ for all $k \in \N$.
\end{enumerate}
\end{theorem}
\begin{theorem}[\textbf{Divergence Criteria}]
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}
\begin{theorem}[\textbf{Monotone Subsequence Theorem}]
If $X=(x_n)$ is a sequence of real numbers, then there is a subsequence of $X$ that is monotone.
\end{theorem}
\begin{theorem}[\textbf{The Bolzano-Wierstrass Theorem}]
A bounded sequence of real numbers has a convergent subsequence.
\end{theorem}
\begin{theorem}
Let $X=(x_n)$ be a bounded sequence of real numbers and let $x \in \R$ have the property that every convergent subsequence of $X$ converges to $x$. Then the sequence $X$ converges to $x$.
\end{theorem}
\begin{definition}
Let $X=(x_n)$ be a bounded sequence of real numbers.
\begin{enumerate}
\item The \textbf{limit superior} of $(x_n)$ is the infimum of the set $V$ of $v \in \R$ such that $v < x_n$ for at most a finite number of $n \in \N$. It is denoted by
\[\lim \sup (x_n)\ \ \ \text{or}\ \ \ \lim \sup X\ \ \ \text{or}\ \ \ \overline{\lim} (x_n)\]
\item The \textbf{limit inferior} of $(x_n)$ is the supremum of the set of $w \in \R$ such that $x_m < w$ for at most a finite number of $m \in \N$. It is denoted by
\[\lim \inf (x_n)\ \ \ \text{or}\ \ \ \lim \inf X\ \ \ \text{or}\ \ \ \overline{\lim}(x_n)\]
\end{enumerate}
\end{definition}
\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}
\begin{theorem}
A bounded sequence $(x_n)$ is convergent if and only if $\lim \sup (x_n)=\lim \inf (x_n)$.
\end{theorem}
chapter-3/the-cauchy-criterion.tex
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\section{The Cauchy Criterion}
\begin{theorem}
A sequence $X=(x_n)$ of real numbers is said to be a \textbf{Cauchy sequence} if for every $\varepsilon >0$ there exists a natural number $H(\varepsilon)$ such that for all natural numbers $n,m \geq H(\varepsilon)$, the terms $x_n, x_m$ satisfy $|x_n-x_m| < \varepsilon$.
\end{theorem}
\begin{lemma}
If $X=(x_n)$ is a convergent sequence of real numbers, then $X$ is a Cauchy sequence.
\end{lemma}
\begin{lemma}
A Cauchy sequence of real numbers is bounded.
\end{lemma}
\begin{theorem}[\textbf{Cauchy Convergence Criterion}]
A sequence of real numbers is convergent if and only if it is a Cauchy sequence.
\end{theorem}
\begin{definition}
We say that a sequence $X=(x_n)$ of real numbers is \textbf{contractive} if there exists a constant $C$, $0<C,1$, such that
\[|x_{n+2}-x_{n+1}| \leq C|x_{n+1}-x_n|\]
for all $n \in \N$. The number $C$ is called the \textbf{constant} of the contractive sequence.
\end{definition}
\begin{theorem}
Every contractive sequence is a Cauchy sequence, and therefore is convergent.
\end{theorem}
\begin{corollary}
If $X:=(x_n)$ is a contractive sequence with constant $C, 0<C<1$, and if $x^*:= \lim X$, then
\begin{enumerate}
\item $|x^*-x_n| \leq \frac{C^{n-1}}{1-C}|x_2-x_1|$,
\item $|x^*-x_n| \leq \frac{C}{1-C}|x_n-x_{n-1}|$.
\end{enumerate}
\end{corollary}