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 12

Definition

Let $\left|a_n\right|$ be a sequence in $ℝ$ or $ℂ\text{.}$

The series $\sum _{n=1}^{\infty }{a}_n$ is the sequence $(s_1,s_2,s_3,\dots ),$ where $s_k=a_1+a_2+\cdots +a_k\text{.}$

The series $\sum _{n=1}^{\infty }{a}_n$ converges to $L$ if the sequence $(s_1,s_2,s_3,\dots )$ converges to $L\text{.}$

Write $\sum _{n=1}^{\infty }{a}_n=L$ if $\sum _{n=1}^{\infty }{a}_n$ converges to $L\text{.}$

Harmonic series and the Riemann zeta function

$$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}& =& {\displaystyle 1+\frac{1}{2}+\underbrace{1/3+\frac{1}{4}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+\cdots }\\ & >& {\displaystyle 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots }\end{array}$$ So $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges. $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}& =& {\displaystyle 1+\underbrace{1/2^2+\frac{1}{3^2}}+\underbrace{\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}}+\underbrace{\frac{1}{8^2}+\cdots }}\\ & <& {\displaystyle 1+\frac{2}{2^2}+\frac{4}{4^2}+\frac{8}{8^2}+\cdots }\\ & =& {\displaystyle 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots }\\ & =& {\displaystyle 1+\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^4+\cdots =\frac{1}{1-\frac{1}{2}}=2\text{.}}\end{array}$$ In fact, according to Wolfram alpha, $$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}\text{.}$$ If $k>1$ then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^k}}& =& {\displaystyle 1+\underbrace{1/2^k+\frac{1}{3^k}}+\underbrace{\frac{1}{4^k}+\frac{1}{5^k}+\frac{1}{6^k}+\frac{1}{7^k}}+\underbrace{\frac{1}{8^k}+\cdots }}\\ & <& {\displaystyle 1+\frac{2}{2^k}+\frac{4}{4^k}+\frac{8}{8^k}+\cdots }\\ & =& {\displaystyle 1+\frac{1}{2^{k-1}}+\frac{1}{4^{k-1}}+\frac{1}{8^{k-1}}+\cdots }\\ & =& {\displaystyle 1+\frac{1}{2^{k-1}}+{\left(\frac{1}{2^{k-1}}\right)}^2+{\left(\frac{1}{2^{k-1}}\right)}^3\cdots }\\ & =& {\displaystyle \frac{1}{1-\frac{1}{2^{k-1}}}=\frac{2^{k-1}}{2^{k-1}-1}}\end{array}$$ so that $\sum _{n=1}^{\infty }\frac{1}{n^k}$ converges.

If $k<1$ then $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^k}}& =& {\displaystyle 1+\frac{1}{2^k}+\frac{1}{3^k}+\frac{1}{4^k}+\cdots }\\ & >& {\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots }\end{array}$$ so that $\sum _{n=1}^{\infty }\frac{1}{n^k}$ diverges.

Let $k\in {ℝ}_{>0}\text{.}$ $$\sum _{n=1}^{\infty }\frac{1}{n^k}$$ converges if $k>1$ and $$\sum _{n=1}^{\infty }\frac{1}{n^k}$$ diverges if $k\le 1\text{.}$

Let $s\in ℂ\text{.}$ The Riemann zeta function at $s$ is $$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{n^s}\text{.}$$

$\zeta \left(2\right)=\frac{{\pi }^2}{6}\text{.}$

Integral tests

The fundamental theorem of calculus (FTC) says: Let $${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)\text{where}\frac{dF}{dx}=f\text{.}$$ Then $${\int }_a^{b}f\left(x\right)dx=(\text{area under}\hspace{0.17em}y=f\left(x\right)\hspace{0.17em}\text{between}\hspace{0.17em}x=a\hspace{0.17em}\text{and}\hspace{0.17em}x=b)$$ if you can make good sense of what "area under $y=f\left(x\right)$ between $x=a$ and $x=b\text{"}$ means. We want to use FTC to think about series.

$a_n=\frac{1}{n}$ and $\sum _{n=1}^{\infty }\frac{1}{n}\text{.}$

If $F\left(x\right)=\text{log}\left(x\right)$ then $\frac{dF}{dx}=\frac{1}{x}$ and $$\begin{array}{ccc}{\displaystyle {\int }_a^b\frac{1}{x}dx}& =& {\displaystyle \text{log}\left(b\right)-\text{log}\left(a\right)}\\ & =& {\displaystyle (\text{area under}\hspace{0.17em}y=\frac{1}{x}\hspace{0.17em}\text{between}\hspace{0.17em}x=a\hspace{0.17em}\text{and}\hspace{0.17em}x=b)}\end{array}$$ $$\begin{array}{c} y=1x 1 2 3 4 5 x 1 1 2 1 5 y \end{array}$$ So $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}& =& {\displaystyle 1+\frac{1}{2}+1/3+\frac{1}{4}+\cdots }\\ & =& {\displaystyle \text{area of the shaded boxes}}\\ & >& {\displaystyle \text{area under}\hspace{0.17em}y=\frac{1}{x}\hspace{0.17em}\text{from}\hspace{0.17em}x=1\hspace{0.17em}\text{to}\hspace{0.17em}x=100000000000}\\ & =& {\displaystyle \text{log}\left(100000000000\right)-\text{log}\left(1\right)}\\ & =& {\displaystyle \text{VERY LARGE.}}\end{array}$$ So $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges.

$\left(a_n\right)=\frac{1}{n^2}$ and $\sum _{n=1}^{\infty }\frac{1}{n^2}\text{.}$

If $F\left(x\right)=-\frac{1}{x}$ then $\frac{dF}{dx}=\frac{1}{x^2}$ and $${\int }_a^b\frac{1}{x^2}dx=-\frac{1}{b}-(-\frac{1}{a})=(\text{area under}\hspace{0.17em}y=\frac{1}{x^2}\hspace{0.17em}\text{between}\hspace{0.17em}x=a\hspace{0.17em}\text{and}\hspace{0.17em}x=b)\text{.}$$ $$\begin{array}{c} y=1x2 1 2 3 x 1 1 4 y \end{array}$$ Then So $\sum _{n=1}^{\infty }\frac{1}{n^2}<2\text{.}$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100326suggLect12.pdf and was given on 26 March 2010.

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