# Monotone Sequences

## Definition 2.3.1

A sequence $\{a_n\}$ is said to be
(a) monotone increasing if $a_n \leq a_{n+1}$ for all $n\in\mathbb{N}$
(b) monotone decreasing if $a_n \geq a_{n+1}$ for all $n\in\mathbb{N}$
(c) monotone if it is either monotone increasing or monotone decreasing

## Theorem 2.3.2 (Monotone Convergence Theorem)

If a sequence $\{a_n\}$ is monotone and bounded, then it converges.

<proof> Suppose that $\{a_n\}$ is bounded and monotone increasing. Put \begin{align} E:=\{a_n:n\in\mathbb{N}\} \neq \emptyset \end{align} Then $E$ is bounded above and nonempty set. By the least upper bound property, we have $\alpha=\sup E$. Now, we want to show that \begin{align} \lim_{n\to\infty}a_n = \alpha \end{align} Let $\epsilon >0$ be given. Then there exists $n_0\in\mathbb{N}$ s.t. $\alpha-\epsilon <a_{n_0}\leq \alpha$. Since $\{a_n\}$ is increasing, $a_n\geq a_{n_0}$ for all $n\geq n_0$. Thus, if $n\geq n_0$, then \begin{align} \alpha-\epsilon < a_n \leq \alpha<\alpha+\epsilon \end{align} $\therefore \displaystyle{\lim_{n\to\infty}a_n=\alpha}$ $\tag*{\square}$

## Corollary 2.3.3 (Nested interval property)

Let $\{I_n\}$ be a sequence of closed and bounded intervals with $I_n\supset I_{n+1}$ for all $n\in\mathbb{N}$.
$\Rightarrow \cap_{n=1}^\infty I_n\neq\emptyset$

<proof> Let $I_n = [a_n, b_n]$. Since $I_n \supset I_{n+m}$ for all $m\geq 0$, \begin{align} a_n \leq a_{n+m} \leq b_{n+m} \leq b_m \end{align} Thus, $a_n \leq b_m$ for all $m\in\mathbb{N}$. Since $\{a_n\}$ is monotone increasing, by the monotone convergence theorem, we know that $\{a_n\}$ converges to $a=\sup\{a_n: n\in\mathbb{N}\}$. Now, note that $a\leq b_m$ for all $m\in\mathbb{N}$. So, we have $a\in I_m$ for all $m\in\mathbb{N}$.

$\therefore a\in \bigcap_{n=1}^\infty I_m$ $\tag*{\square}$

## Examples 2.3.5

$e:= \displaystyle{\lim_{n\to\infty}}(1+\frac{1}{n})^n$

<proof> Define $t_n:= (1+\frac{1}{n})^n$. Then, \begin{align} \begin{split} (1+\frac{1}{n})^n&=1 + {n \choose 1}\frac{1}{n} + {n \choose 2}\frac{1}{n^2} + \cdots {n \choose n-1}\frac{1}{n^{n-1}} + {n \choose n}\frac{1}{n^n} \\ &= 1 + 1 + \frac{n(n-1)}{2}\frac{1}{n^2}+\cdots + \frac{n(n-1)\cdots 2\cdot1}{1\cdot2\cdots n}\frac{1}{n^n} \end{split} \end{align} $(k+1)^{th}$ term of $t_n$ is \begin{align} \frac{n(n-1)\cdots (n-k+1)}{1\cdot 2\cdots k}\frac{1}{n^k}=\frac{1}{k!}(1-\frac{1}{n})\cdots (1-\frac{k-1}{n}) \end{align} Then $t_{n+1}$ is
\begin{align} (1+\frac{1}{n+1})^{n+1} =1 + 1 + \frac{(n+1)n}{2}\frac{1}{(n+1)^2} + \cdots \frac{(n+1)n\cdots 1}{1\cdot 2\cdots (n+1)}\frac{1}{(n+1)^{n+1}} \end{align} $(k+1)^{th}$ term of $t_{n+1}$ is \begin{align} \frac{1}{k!}(1-\frac{1}{n+1})\cdots (1-\frac{k-1}{n+1}) \end{align} Since $(k+1)^{th}$ of $t_n$ is less than that of $t_{n+1}$ for all $k$, we have $t_n<t_{n+1}$ for all $n\in\mathbb{N}$. Moreover, \begin{align} \begin{split} (1+\frac{1}{n})^n &=1 + 1 + \frac{n(n-1)}{2}\frac{1}{n^2}+\cdots + \frac{n(n-1)\cdots 2\cdot1}{1\cdot2\cdots n}\frac{1}{n^n}\\ &<1+1 + \frac{1}{2!} + \cdots + \frac{1}{n!} \\ &\leq 1 = 1 +\frac{1}{2} +\frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} \\ &=1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}} \\ &\leq 3 \end{split} \end{align} By the monotone convergence theorem, $\{t_n\}$ converges. $\tag*{\square}$

## Definition 2.3.6

\begin{align} \lim_{n\to\infty}a_n =\infty &\iff \forall M>0, \exists N\in\mathbb{N} \text{ s.t. } n\geq N \Rightarrow a_n > M \\ \lim_{n\to\infty}a_n =-\infty &\iff \forall M<0, \exists N\in\mathbb{N} \text{ s.t. } n\geq N \Rightarrow a_n < M \\ \end{align}

## Definition 2.4.1

Let $\{p_n\}$ be a sequence and let $\{n_k\}$ be strictly increasing sequence, i.e. $n_1 < n_2 <n_3 < \cdots$. We call $\{p_{n_k}\}_{k=1}^\infty$ a subsequence of $\{p_n\}$.

## Theorem 2.4.3

Let $\{p_n\}$ be a convergent sequence with $\displaystyle{\lim_{n\to\infty}p_n=p}$. Then for all subsequences of $\{p_n\}$ converges to $p$.

<proof> Let $\epsilon >0$ be given and let $\{p_{n_k}\}$ be a subsequence of $\{p_n\}$. Since $p_n\to p$ as $n\to\infty$, there exists $N\in\mathbb{N}$ such that \begin{align} n\geq N \Rightarrow |p_n-p| <\epsilon \end{align} Now, $n_k \geq k$. So, if $k\geq N$, then $n_k\geq k \geq N$.
$\therefore |p_{n_k}-p|<\epsilon$
$\therefore \displaystyle{\lim_{k\to\infty}}p_{n_k}=p$ $\tag*{\square}$

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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