Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 July 2014

Lecture 17

Let $\left(a_n\right)$ be a sequence in $ℝ\text{.}$

(a)	Assume $\underset{n\to \infty }{\text{lim}}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=a$ exists and $a<1\text{.}$ Then $\sum _{n=1}^{\infty }\left|a_n\right|$ converges.
(b)	Assume $\underset{n\to \infty }{\text{lim}}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=a$ exists and $a>1\text{.}$ Then $\sum _{n=1}^{\infty }\left|a_n\right|$ diverges.

Let $\left(a_n\right)$ be a sequence in $ℝ\text{.}$

(a)	Assume $\underset{n\to \infty }{\text{lim}}{\left|a_n\right|}^{\frac{1}{n}}=a$ exists and $a<1\text{.}$ Then $\sum _{n=1}^{\infty }\left|a_n\right|$ converges.
(b)	Assume $\underset{n\to \infty }{\text{lim}}{\left|a_n\right|}^{\frac{1}{n}}=a$ exists and $a>1\text{.}$ Then $\sum _{n=1}^{\infty }\left|a_n\right|$ diverges.

		Proof of theorem 1a.
	Assume $\underset{n\to \infty }{\text{lim}}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=a$ and $a<1\text{.}$ Let $\epsilon \in {ℝ}_{>0}$ such that $a<a+\epsilon <1\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ such that if $n>N$ then $$|\frac{\left|a_{n+1}\right|}{\left|a_n\right|}-a|<\epsilon \text{.}$$ Then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}& =& {\displaystyle \left|a_1\right|+\left|a_2\right|+\cdots +\left|a_N\right|+\left|a_{N+1}\right|+\left|a_{N+2}\right|+\cdots }\\ & =& {\displaystyle \left|a_1\right|+\left|a_2\right|+\cdots +\left|a_N\right|+\left|a_N\right|·\frac{\left|a_{N+1}\right|}{\left|a_N\right|}+\left|a_N\right|\frac{\left|a_{N+1}\right|}{\left|a_N\right|}+\frac{\left|a_{N+2}\right|}{\left|a_{N+1}\right|}+\cdots }\\ & \le & {\displaystyle \left|a_1\right|+\left|a_2\right|+\cdots +\left|a_N\right|+\left|a_N\right|(\frac{\left|a_{N+1}\right|}{\left|a_N\right|}+\frac{\left|a_{N+1}\right|}{\left|a_N\right|}\frac{\left|a_{N+2}\right|}{\left|a_{N+1}\right|}+\frac{\left|a_{N+1}\right|}{\left|a_N\right|}\frac{\left|a_{N+2}\right|}{\left|a_{N+1}\right|}\frac{\left|a_{N+3}\right|}{\left|a_{N+2}\right|}+\cdots )}\\ & \le & {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+\left|a_N\right|((a+\epsilon )+{(a+\epsilon )}^2+{(a+\epsilon )}^3+\cdots )}\\ & =& {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+\left|a_N\right|\left(\frac{1}{1-(a-\epsilon )}\right)\text{.}}\end{array}$$ So the ratio test is a comparison to a geometric series! $\square$

		Proof of theorem 2a.
	Assume $\underset{n\to \infty }{\text{lim}}{\left|a_n\right|}^{\frac{1}{n}}=a$ and $a<1\text{.}$ Let $\epsilon \in {ℝ}_{>0}$ such that $a<a+\epsilon <1\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ such that if $n>N$ then $|{\left|a_n\right|}^{\frac{1}{n}}-a|<\epsilon \text{.}$ Then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}& =& {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+\left|a_{N+1}\right|+\left|a_{N+2}\right|+\cdots }\\ & =& {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+{\left({\left|a_{N+1}\right|}^{\frac{1}{N+1}}\right)}^{N+1}+{\left({\left|a_{N+2}\right|}^{\frac{1}{N+2}}\right)}^{N+2}+\cdots }\\ & \le & {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+{(a+\epsilon )}^{N+1}+{(a+\epsilon )}^{N+2}+\cdots }\\ & =& {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+{(a+\epsilon )}^{N+1}(1+(a+\epsilon )+{(a+\epsilon )}^2+\cdots )}\\ & =& {\displaystyle \left|a_1\right|+\cdots +\left|a_N\right|+{(a+\epsilon )}^{N+1}\left(\frac{1}{1-(a+\epsilon )}\right)\text{.}}\end{array}$$ $\square$

A sequence $\left(a_n\right)$ is contractive if there exists $\alpha \in ℝ,$ $\alpha \in (0,1)$ such that $$|a_{n+1}-a_n|\le \alpha |a_n-a_{n-1}|\text{for}\hspace{0.17em}n=2,3,4,\dots$$ If $\left(a_n\right)$ is contractive then $$\begin{array}{ccc}{\displaystyle |a_{n+1}-a_n|}& \le & {\displaystyle \alpha |a_n-a_{n-1}|}\\ & \le & {\displaystyle {\alpha }^2|a_{n-1}-a_{n-2}|}\\ & \le & {\displaystyle {\alpha }^3|a_{n-2}-a_{n-3}|}\\ & \le & {\displaystyle ⋮}\\ & \le & {\displaystyle {\alpha }^{n-1}\mid a_{n(n-2)}-a_{n-(n-1)}\mid }\\ & \le & {\displaystyle {\alpha }^{n-1}|a_2-a_1|,}\end{array}$$ which is very small if $n=10000000$ and $\alpha =\frac{1}{2}\text{.}$ This is the idea behind the proof of the ratio test.

A sequence $\left(a_n\right)$ is Cauchy if $\left(a_n\right)$ satisfies: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m,n>N$ then $$d(a_m,a_n)<\epsilon \text{.}$$

There does not exist $\frac{a}{b}\in \mathbb{Q}$ such that ${\left(\frac{a}{b}\right)}^2=2\text{.}$

		Proof.
	Proof by contradiction. Assume $\frac{a}{b}\in \mathbb{Q}$ and $\frac{a^2}{b^2}=2$ and $\frac{a}{b}$ is reduced. Then $a^2=2b^2,$ so that $a^2$ is even. So $a$ is even. So $2b^2$ is divisible by $4\text{.}$ So $b^2$ is even. So $b$ is even. So $\frac{a}{b}$ is not reduced. Contradiction. So there does not exist $\frac{a}{b}\in \mathbb{Q}$ with ${\left(\frac{a}{b}\right)}^2=2\text{.}$ $\square$

Consider the sequence in $\mathbb{Q}\text{:}$ $$(1,\frac{14}{10},\frac{141}{100},\frac{1414}{1000},\frac{14142}{10000},\frac{141421}{100000},\dots )$$ This is a Cauchy sequence that does not converge.

Consider the sequence in $ℝ\text{:}$ $$(1.00\dots ,1.4000,\dots ,1.4100\dots ,1.41400\dots ,1.414200,\dots )$$ This is a Cauchy sequence that does converge.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100416Lect17.pdf and was given on 16 April 2010.

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