Real Analysis

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

Last update: 12 July 2014

Lecture 13

Conditional convergence and alternating series

Let $a_n$ be a sequence in ???. If $\sum _{n=1}^{\infty }\left|a_n\right|$ converges then $\sum _{n=1}^{\infty }{a}_n$ converges.

A series $\sum _{n=1}^{\infty }{a}_n$ is conditionally convergent if $\sum _{n=1}^{\infty }{a}_n$ converges but $\sum _{n=1}^{\infty }\left|a_n\right|$ diverges.

A series $\sum _{n=1}^{\infty }{a}_n$ is absolutely convergent if $\sum _{n=1}^{\infty }\left|a_n\right|$ converges.

(Favourite example) Does $\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{1}{n}$ converge? $$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{1}{n}=1-\frac{1}{2}+1/3-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$ Since $$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$ then $$\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots$$ and $$\text{log}\left(1+x\right)=\int \frac{1}{1+x}dx=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots$$ So if $1-\frac{1}{2}+1/3-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$ converges it is equal to $\text{log}(1+1)=\text{log}\hspace{0.17em}2\text{.}$ A quick experiment indicates that it does converge. However $$\sum _{n=1}^{\infty }\left|{(-1)}^{n-1}\frac{1}{n}\right|=\sum _{n=1}^{\infty }\frac{1}{n}\text{diverges (harmonic series).}$$ So $\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{1}{n}$ is conditionally convergent and not absolutely convergent.

(Liebniz's Theorem) If $\left(a_n\right)$ is a sequence in $ℝ$ such that

(a)	$a_n\in {ℝ}_{\ge 0},$
(b)	if $n\in {\mathbb{Z}}_{>0}$ then $a_n\ge a_{n+1},$
(c)	$\underset{n\to \infty }{\text{lim}}{a}_n=0$

then $\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}{a}_n$ converges.

		Proof.
	Assume $\left(a_n\right)$ is a sequence in $ℝ$ and $a_n\in {ℝ}_{\ge 0},$ and if $n\in {\mathbb{Z}}_{>0}$ then $a_n\ge a_{n+1}$ and $\underset{n\to \infty }{\text{lim}}{a}_n=0\text{.}$ To show: $\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}{a}_n=a_1-a_2+a_3-a_4+a_5-\cdots$ converges. Let $$s_{2m}=(a_1-a_2)+(a_3-a_4)+\cdots +(a_{2m-1}-a_{2m})\text{.}$$ Then $s_{2m}\le s_{2(m+1)}\text{.}$ Since $s_{2m}=a_1-(a_2-a_3)-(a_4-a_5)-\cdots -(a_{2m-2}-a_{2m-1})-a_{2m},$ then $s_{2m}\le a_1\text{.}$ So the sequence $(s_2,s_4,s_6,\dots )$ is increasing and bounded above. So $\underset{n\to \infty }{\text{lim}}{s}_{2m}$ exists. Let $\ell =\underset{m\to \infty }{\text{lim}}{s}_{2m}\text{.}$ Since $s_{2m+1}=s_{2m}+a_{2m+1}$ then $$\begin{array}{ccc}{\displaystyle \underset{m\to \infty }{\text{lim}}{s}_{2m+1}}& =& {\displaystyle \underset{m\to \infty }{\text{lim}}{s}_{2m}+\underset{m\to \infty }{\text{lim}}{a}_{2m+1}}\\ & =& {\displaystyle \ell +0=\ell \text{.}}\end{array}$$ So $\underset{m\to \infty }{\text{lim}}{s}_m=\ell \text{.}$ $\square$

$\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{\text{log}\hspace{0.17em}n}{\sqrt{n}}\text{.}$

Since $$\begin{array}{ccc}{\displaystyle \frac{d}{dx}\left(\frac{\text{log}\hspace{0.17em}x}{\sqrt{x}}\right)}& =& {\displaystyle \frac{d}{dx}\left(x^{-\frac{1}{2}}\text{log}\hspace{0.17em}x\right)}\\ & =& {\displaystyle x^{-\frac{1}{2}}\frac{1}{x}+-\frac{1}{2}{x}^{-\frac{3}{2}}\text{log}\hspace{0.17em}x}\\ & =& {\displaystyle x^{-\frac{3}{2}}(1-\frac{1}{2}\text{log}\hspace{0.17em}x)}\end{array}$$ is less than $0$ for $x>e^2,$ the sequence $\frac{\text{log}\hspace{0.17em}n}{\sqrt{n}}$ is decreasing for $x>9\text{.}$

Since $\underset{n\to \infty }{\text{lim}}\frac{\text{log}\hspace{0.17em}n}{n^{\frac{1}{2}}}=\underset{n\to \infty }{\text{lim}}{e}^{-\frac{1}{2}\text{log}\hspace{0.17em}n}\text{log}\hspace{0.17em}n=\underset{y\to \infty }{\text{lim}}{e}^{-\frac{1}{2}y}y=0$ Leibniz theorem gives that $\sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{\text{log}\hspace{0.17em}n}{\sqrt{n}}$ converges.

Thinking about rearrangements

$$1-\frac{1}{2}+1/3-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots =\text{log}\hspace{0.17em}2\text{.}$$ $$\begin{array}{ccc}{\displaystyle -\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\cdots }& =& {\displaystyle -\frac{1}{2}(1+\frac{1}{2}+1/3+\frac{1}{4}+\cdots )}\\ & =& {\displaystyle -\frac{1}{2}\left(\text{VERY VERY LARGE}\right)}\end{array}$$ $$1+1/3+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots >\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots =\text{VERY VERY LARGE}$$ Pick a number $\ell \in {ℝ}_{>0}\text{.}$

If we take $$1+1/3+\frac{1}{5}+\frac{1}{7}+\cdots +\frac{1}{299}\text{just until we get larger than}\hspace{0.17em}\ell ,$$ then add $$-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\cdots -\frac{1}{70}\text{just until we get smaller than}\hspace{0.17em}\ell ,$$ then add $$\frac{1}{301}+\frac{1}{303}+\cdots \text{just until we get larger than}\hspace{0.17em}\ell \hspace{0.17em}\text{again,}$$ then add $$-\frac{1}{72}-\frac{1}{74}-\cdots \text{just until we get smaller than}\hspace{0.17em}\ell \hspace{0.17em}\text{again}$$ $$⋮$$ This process will create a series that converges to $\ell \text{.}$

Assume that $a_n$ is a sequence in $ℝ$ or $ℂ$ If $\sum _{n=1}^{\infty }\left|a_n\right|$ converges then $\sum _{n=1}^{\infty }{a}_n$ converges.

		Proof.
	Let $A_n=\left|a_1\right|+\left|a_2\right|+\cdots +\left|a_n\right|$ and $s_n=a_1+a_2+\cdots +a_n\text{.}$ Since $\sum _{n=1}^{\infty }\left|a_n\right|=(A_1,A_2,A_3,\dots )$ converges the sequence $A_n$ is Cauchy. Since $$\begin{array}{ccc}{\displaystyle |s_m-s_n|}& =& {\displaystyle |a_{n+1}+\cdots +a_m|}\\ & \le & {\displaystyle \left|a_{m+1}\right|+\cdots +\left|a_n\right|=|A_m-A_n|,}\end{array}$$ the sequence $s_n$ is Cauchy. Since Cauchy sequences converge in $ℝ$ or $ℂ$ (or any complete metric space) $$\sum _{n=1}^{\infty }{a}_n=\left(s_n\right)$$ converges. $\square$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100329suggLect13.pdf and was given on 29 March 2010.

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