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 14

The Ratio test

Does $\sum _{n=1}^{\infty }\frac{n!}{n^n}$ converge? $$\sum _{n=1}^{\infty }\frac{n!}{n^n}=1+\frac{2}{2^2}+\frac{3!}{3^3}+\frac{4!}{4^4}+\cdots$$ Look at $$\begin{array}{ccc}{\displaystyle \mid \frac{\frac{(n+1)!}{{(n+1)}^{n+1}}}{\frac{n!}{n^n}}\mid }& =& {\displaystyle \mid \frac{(n+1)!n^n}{n!{(n+1)}^{n+1}}\mid =\mid \frac{(n+1)n^n}{{(n+1)}^{n+1}}\mid }\\ & =& {\displaystyle \frac{n^n}{{(n+1)}^n}={\left(\frac{n}{n+1}\right)}^n=\frac{1}{{(1+\frac{1}{n})}^n}}\\ & <& {\displaystyle \frac{1}{2}\text{if}\hspace{0.17em}n\hspace{0.17em}\text{is large enough.}}\end{array}$$ In fact $\frac{1}{{(1+\frac{1}{1})}^1}=\frac{1}{2}$ and $\frac{1}{{(1+\frac{1}{2})}^2}=\frac{1}{{\left(\frac{3}{2}\right)}^2}=\frac{4}{9}<\frac{1}{2}\text{.}$ So $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{n!}{n^n}}& =& {\displaystyle 1+\frac{2!}{2^2}+\frac{3!}{3^3}+\frac{4!}{4^4}+\cdots }\\ & =& {\displaystyle 1+\frac{2!}{2^2}+\frac{2!}{2^2}\left(\frac{3!}{3^3}\frac{2!}{2^2}\right)+\frac{2!}{2^2}\left(\frac{3!}{3^3}\frac{2!}{2^2}\right)\left(\frac{4!}{4^4}\frac{3!}{3^3}\right)+\cdots }\\ & <& {\displaystyle 1+\frac{2!}{2^2}+\frac{2!}{2^2}·\frac{1}{2}+\frac{2!}{2^2}·\frac{1}{2}·\frac{1}{2}+\cdots }\\ & =& {\displaystyle 1+\frac{2!}{2^2}(1+\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\cdots )}\\ & =& {\displaystyle 1+\frac{2!}{2^2}\left(\frac{1}{1-\frac{1}{2}}\right)=1+\frac{2}{4}·2=1+1=2\text{.}}\end{array}$$ So $\sum _{n=1}^{\infty }\frac{n!}{n^n}$ converges.

Let $\left(a_n\right)$ be a sequence in ${ℝ}_{\ge 0}\text{.}$

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

		Proof.
	Assume $\underset{n\to \infty }{\text{lim}}\frac{a_{n+1}}{a_n}=a$ exists and $a<1\text{.}$ Let $\epsilon \in {ℝ}_{>0}$ so that $a+\epsilon <1\text{.}$ Since $\underset{n\to \infty }{\text{lim}}\frac{a_{n+1}}{a_n}=a$ there exists $N\in {\mathbb{Z}}_{>0}$ with $\frac{a_{n+1}}{a_n}<a+\epsilon$ if $n\in {\mathbb{Z}}_{>0}$ with $n>N\text{.}$ Then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }{a}_n}& =& {\displaystyle a_1+a_2+\cdots +a_N+a_{N+1}+a_{N+2}+\cdots }\\ & =& {\displaystyle a_1+\cdots +a_N+a_{N+1}+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)+\cdots }\\ & <& {\displaystyle a_1+\cdots +a_N+a_{N+1}+a_{N+1}(a+\epsilon )+a_{N+1}{(a+\epsilon )}^2+\cdots }\\ & =& {\displaystyle a_1+\cdots +a_N+a_{N+1}+(1+(a+\epsilon )+{(a+\epsilon )}^2+{(a+\epsilon )}^3+\cdots )}\\ & =& {\displaystyle a_1+\cdots +a_N+a_{N+1}\left(\frac{1}{1-(a+\epsilon )}\right)\text{.}}\end{array}$$ So $\sum _{n=1}^{\infty }{a}_n$ converges. (b) Assume $\underset{n\to \infty }{\text{lim}}\frac{a_{n+1}}{a_n}$ exists and $a>1\text{.}$ Let $\epsilon \in {ℝ}_{>0}$ such that $a-\epsilon >1\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $\frac{a_{n+1}}{a_n}>a-\epsilon \text{.}$ Then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }{a}_n}& =& {\displaystyle a_1+\cdots +a_N+a_{N+1}+a_{N+2}+a_{N+3}+\cdots }\\ & =& {\displaystyle a_1+\cdots +a_N+a_{N+1}(1+\frac{a_{N+2}}{a_{N+1}}+\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)+\cdots )}\\ & >& {\displaystyle a_1+\cdots +a_N+a_{N+1}(1+(a-\epsilon )+{(a-\epsilon )}^2+\cdots )}\\ & >& {\displaystyle a_1+\cdots +a_N+a_{N+1}(1+1+1+1+\cdots )}\end{array}$$ So $\sum _{n=1}^{\infty }{a}_n$ diverges. $\square$

Radius of convergence

Find the radius of convergence and interval of convergence of $$\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{\sqrt[3]{n}}$$ i.e. for which $x$ does the series converge?

To analyse: $$\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{n^{1/3}}=\sum _{n=1}^{\infty }\frac{y^n}{n^{1/3}},$$ if $y=2x-1\text{.}$

Look at the ratios: $$\mid \frac{\frac{y^{n+1}}{{(n+1)}^{1/3}}}{\frac{y^n}{n^{1/3}}}\mid =\frac{\left|y\right|n^{1/3}}{{(n+1)}^{1/3}}=\frac{\left|y\right|}{{(1+\frac{1}{n})}^{1/3}}$$ As $n\to \infty ,$ $$\underset{n\to \infty }{\text{lim}}\frac{\left|y\right|}{{(1+\frac{1}{n})}^{1/3}}=\frac{\left|y\right|}{{(1+0)}^{1/3}}=\left|y\right|\text{.}$$ So, if $\left|y\right|<1$ then $\sum _{n=1}^{\infty }\left|\frac{y^n}{n^{1/3}}\right|$ converges and if $\left|y\right|>1$ then $\sum _{n=1}^{\infty }\left|\frac{y^n}{n^{1/3}}\right|$ diverges.

If $\left|y\right|=1$ then $$\sum _{n=1}^{\infty }\left|\frac{y^n}{n^{1/3}}\right|=\sum _{n=1}^{\infty }\frac{1}{n^{1/3}}>\sum _{n=1}^{\infty }\frac{1}{n}$$ diverges.

So, we can be sure that if $\left|y\right|<1$ then $\sum _{n=1}^{\infty }\frac{y^n}{n^{1/3}}$ converges.

So if $|2x-1|<1$ then $\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{n^{1/3}}$ converges.

So, if $|x-\frac{1}{2}|<\frac{1}{2}$ then $\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{n^{1/3}}$ converges.

So, if $x$ is inside the circle $$\begin{array}{c} 1 2 + 0 i 1 + 0 i not i axis - i - 1 2 i 0 + i 1 2 i i axis \end{array}$$ then $\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{n^{1/3}}$ converges.

Let $r,s\in ℂ\text{.}$ Assume $\sum _{n=0}^{\infty }{a}_n{s}^n$ converges. If $\left|r\right|<\left|s\right|$ then $\sum _{n=0}^{\infty }{a}_n{\left|r\right|}^n$ converges.

		Proof.
	Since $\sum _{n=0}^{\infty }{a}_n{s}^n$ converges $\underset{n\to \infty }{\text{lim}}\left|a_n{s}^n\right|=0\text{.}$ Let $\epsilon \in {ℝ}_{>0}\text{.}$ Then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $\left|a_n{s}^n\right|<\epsilon \text{.}$ Then $$\begin{array}{ccc}{\displaystyle \sum _{n=0}^{\infty }\left|a_n{r}^n\right|}& =& {\displaystyle \left|a_0\right|+\left|a_{1}r\right|+\cdots +\left|a_N{r}^N\right|+\left|a_{N+1}{r}^{N+1}\right|+\cdots }\\ & =& {\displaystyle \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\left|a_{N+1}{s}^{N+1}\right|\left|\frac{r^{N+1}}{s^{N+1}}\right|+\left|a_{N+2}{s}^{N+2}\right|\left|\frac{r^{N+2}}{s^{N+2}}\right|+\cdots }\\ & <& {\displaystyle \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|+\epsilon \left|\frac{r^{N+2}}{s^{N+2}}\right|+\epsilon \left|\frac{r^{N+3}}{s^{N+3}}\right|+\cdots }\\ & =& {\displaystyle \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|(1+\left|\frac{r}{s}\right|+\left|\frac{r^s}{s^2}\right|+\cdots )}\\ & =& {\displaystyle \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|\left(\frac{1}{1-\frac{\left|r\right|}{\left|s\right|}}\right)\text{.}}\end{array}$$ So $\sum _{n=0}^{\infty }\left|a_n\right|\left|r^n\right|$ converges. So $\sum _{n=0}^{\infty }{a}_n{r}^n$ converges. $\square$

It follows that, if $x$ is outside the circle $$\begin{array}{c} 1 2 + 0 i 1 + 0 i not i axis - i - 1 2 i 0 + i 1 2 i i axis \end{array}$$ then $\sum _{n=1}^{\infty }\frac{{(2x-1)}^n}{n^{1/3}}$ diverges.

What if $x$ is on the circle?

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100331suggLect14.pdf and was given on 31 March 2010.

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