Lecture 2 - Sequences

A sequence is a set $\{x_n|n\in \mathbb{N}\}$, denoted by $(x_n{)}_{n\in \mathbb{N}}$ or simply $(x_n)$. A sequence $(x_n)$ is:

• bounded (above or below) if the set $\{x_n|n\in \mathbb{N}\}$ is bounded
• increasing if $x_{n+1}\ge x_n,\forall n\in \mathbb{N}$
• monotone if it is either increasing pr decreasing A sequence $(x_n)$ has a limit $l\in \overline{\mathbb{R}}$, and we write $\underset{n\to \infty }{\lim }=l$ or $x_n\to l$ $\forall V\in \mathcal{V}(l),$
• 🔼 Finish completing this using lecture notes

Proposition

Any convergent sequence is bounded

Proof

Let $(x_n)\text{ convergent to }l\in mathbbR$. Let $\epsilon =1$. Then since $x_n\to l,\exists N_1\in \mathbb{N}$, $|x_n-l|<1,\forall n\ge N_1$ $x_n\in (l-1,l+1)\forall n\ge N_1⟹\text{ bounded (starting from }{N}_1\text{)}$ $M=max|x_n|,n=\overline{1,N_1}$ $(x_n)$ is bounded by max $\{l+1,M\}$

Theorem (Weierstrass)

Any monotone and bounded sequence is convergent

Proof

let $(x_n)$ increasing and bounded. $x_1\le x_2\le \dots \le x_n\le x_{n+1}\le \dots$ $A=\{x_n|n\in \mathbb{N}\}=(x_n)$, bounded $\forall \epsilon >0,\exists N\in \mathbb{N}\text{ s.t. }:sup(A)-\epsilon \le x_N\le sup(A)$ $sup(A)-\epsilon <x_N\le x_n\le sup(A),\forall n\ge N$ $|x_n|-sup(A)<\epsilon ,\forall n\in \mathbb{N}⟹sup(A)=\underset{n\to \infty }{\lim }{x}_n$

Proposition

Any monotone sequence has a limit in $\overline{\mathbb{R}}$.

Proof

$(x_n)$ monotone a. $(x_n)$ is bounded $\stackrel{\text{Weierstrass}}{⟹}(x_n)\text{ is convergent}$ b. $(x_n)$ is unbounded $⟹x_n\to \{\pm \infty \}$

Theorem (Squeeze/Sandwich theorem)

Let $(x_n),(y_n),(z_n)$ be sequences fpr which there exists $n_0\in \mathbb{N}$ such that $x_n\le y_n\le z_n\forall n\ge n_0$ and $\underset{n\to \infty }{\lim }{x}_n=\underset{n\to \infty }{\lim }{z}_n$ Then $\underset{n\to \infty }{\lim }{x}_n=\underset{n\to \infty }{\lim }{y}_n=\underset{n\to \infty }{\lim }{z}_n$

Proof

$x_n\to l:\exists N_1\in \mathbb{N}:|x_n-l|<\epsilon ,\forall n\ge N_1,x_n\in (l-\epsilon ,l+\epsilon )$ $z_n\to l:\exists N_w\in \mathbb{N}:|z_n-l|<\epsilon ,\forall n\ge N_2,z_n\in (l-\epsilon ,l+\epsilon )$ $⟹x_n,z_n\in (l-\epsilon ,l+\epsilon )\forall n\ge max\{N_1,N_2\}=N$ $⟹y_n\in (l-\epsilon ,l+\epsilon ),\forall n\ge N,|y_n-l|<\epsilon$

Theorem (Cantor's nested intervals)

Let $(a_n)$ be increasing and $(b_n)$ be decreasing such that $a_n\le a_{n+1}\le b_{n+1}\le b_n,\forall n\in \mathbb{N}.$ Consider the closed intervals $I_n:=[a_n,b_n],with{I}_{n+1}\subseteq I_n.$ If $\underset{n\to \infty }{\lim }(b_n-a_n)=0$, then there exists $x\in \mathbb{R}$ such that $\bigcap _{n=1}^{\infty }{I}_n=\{x\}$

Proof

Found in lecture notes

Theorem (Bolzano-Weierstrass)

Any bounded sequence has a convergent subsequence

Subsequences

For any sequence $(x_n{)}_{n\in \mathbb{N}}$, subsequences are basically a rule, i.e. $(x_{2n}),(x_{2n+1}),(x_{n^2})$. $(x_i{)}_{i\in I},I\le \mathbb{N}\text{ infinite}$. It is sometimes written as $(x_{n_k}),k\in \mathbb{N}$

Hint for proof: Divide et Impera + Cantor’s nested intervals, $x_{n_1}\in [a_1,b_1],x_{n_2}\in [a_2,b_2],\dots ,x_{n_k}\in [a_k,b_k]$

Proof

$(x_n)$ is bounded $⟹inf(x_n),sup(x_n)\in \mathbb{R}$ Let $a_1=inf(x_n),b_1=sup(x_n)$ $x_n\in [a_1,b_1],\forall n\in \mathbb{N}$ Divide $[a_1,b_1]$ into 2 equal intervals Choose the one that contains an infinity of terms $(x_n)$. $[a_2,b_2],b_2-a_2=\frac{b_1-a_1}{2},[a_2,b_2]\subseteq [a_1,b_1]$ Iteration $[a_k,b_k],b_k-a_k=\frac{b_1-a_1}{2^{k-1}}\to 0,\text{ as }k\to \infty$ $\underset{k\to \infty }{\lim }{x}_{n_k}=\{x\}=\bigcap _{k=1}^{\infty }[a_k,b_k]$

Definition

For a sequence $(x_n)$ we define the set of its limit points by $LIM(x_n):=\{x\in \overline{\mathbb{R}}|\text{ there exists a subsequence }(x_{n_k})\text{ s.t. }{x}_{n_k}\to x\}$ and

\lim_{n\rightarrow\infty} inf\ x_n\coloneqq inf(LIM(x_n)),\\ \lim_{n\rightarrow\infty} sup\ x_n\coloneqq sup(LIM(x_n)),\\ \end{gathered}$$

Example

$x_n=\frac{(-1{)}^{n}n}{n+1},-\frac{1}{2},\frac{2}{3},-\frac{3}{4},\frac{4}{5},\dots$ $x_{2n}=\frac{2n}{2n+1}\to 1,x_{2n+1}=-\frac{2n+1}{2n+2}\to -1$

$LIM(x_n)=\{-1,1\},\underset{n\to \infty }{\lim }inf{x}_n=-1,\underset{n\to \infty }{\lim }sup{x}_n=1$

Definition (Cauchy sequence)

A sequence $(x_n)$ is called Cauchy (or fundamental) if $\forall \epsilon >0,\exists N_{\epsilon }\in \mathbb{R}\text{ such that }|x_m-x_n|<\epsilon ,\forall m,n\ge N_{\epsilon }$

Proposition

Any Cauchy sequence is bounded

Proof

Let $\epsilon =1:\exists N_1\in \mathbb{N}\text{ s.t. }|x_n-x_m|<1,\forall n,m\ge N_1$ In particalular $|x_n-x_{N_1}|<1,\forall n\ge N_1⟹(x_n)\text{ is bounded. }{x}_n\in (x_{N_1}-1,x_{N_1}+1)$

Theorem

A sequence is convergent if and only if it is Cauchy

Proof

$A\iff B:A⟹B,B⟹A$ Convergent $⟹$ Cauchy Let $l=\underset{n\to \infty }{\lim }{x}_n$ $\forall \epsilon >0,\exists N_{\epsilon }\in \mathbb{N}\text{ s.t. }|x_n-l|<\frac{\epsilon }{2},\forall n\ge N_{\epsilon }$ $|x_n-x_m|=|x_n-l+l-x_m|\le \underset{<\frac{\epsilon }{2}}{\underbrace{|x_n-l|}}+\underset{<\frac{\epsilon }{2}}{\underbrace{|x_m-l|}}<\epsilon ,\forall n,m\ge N_{\epsilon }⟹(x_n)\text{ is Cauchy }$