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 16

Examples of improper integrals and series

Analyse ${\int }_1^{\infty }\frac{\text{log}\hspace{0.17em}x}{x}dx\text{.}$

Let $y=\frac{\text{log}\hspace{0.17em}x}{x}=x^{-1}\text{log}\hspace{0.17em}x\text{.}$ Then $$\frac{dy}{dx}=x^{-1}\frac{1}{x}+(-1)x^{-2}\text{log}\hspace{0.17em}x=(1-\text{log}\hspace{0.17em}x)\frac{1}{x^2}\text{.}$$ This is $>0$ for $x<e^1,$ $=0$ for $x=e^1,$ $<0$ for $x>e^1\text{.}$

So $y$ is increasing for $x<e^1,$ has a maximum at $x=e^1\approx \mathrm{2.781828...},$ is decreasing for $x>e^1\text{.}$ $$\begin{array}{c} 1 2 3 e 1 x 1 1 e y \end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_1^{\infty }\frac{\text{log}\hspace{0.17em}x}{x}dx}& =& {\displaystyle \underset{b\to \infty }{\text{lim}}{\int }_1^b\frac{\text{log}\hspace{0.17em}x}{x}dx}\\ & =& {\displaystyle \underset{b\to \infty }{\text{lim}}\left(\frac{{\left(\text{log}\hspace{0.17em}x\right)}^2}{2}{|}_{x=1}^{x=b}\right)}\\ & =& {\displaystyle \underset{b\to \infty }{\text{lim}}(\frac{{\left(\text{log}\hspace{0.17em}b\right)}^2}{2}-\frac{{\left(\text{log}\hspace{0.17em}1\right)}^2}{2})}\\ & =& {\displaystyle \text{divergent.}}\end{array}$$

$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}\text{.}$ $$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}=-0+\frac{\text{log}\hspace{0.17em}2}{2}-\frac{\text{log}\hspace{0.17em}3}{3}+\sum _{n=4}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}\text{.}$$ Since the values $\frac{\text{log}\hspace{0.17em}n}{n}$ are positive, decreasing (see the previous page) and approaching $0$ for $n\ge 4,$ the alternating series test guarantees that $\sum _{n=4}^{\infty }{(-1)}^n\frac{\text{log}\hspace{0.17em}n}{n}$ converges. So $$\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}=\frac{\text{log}\hspace{0.17em}2}{2}-\frac{\text{log}\hspace{0.17em}3}{3}+\sum _{n=4}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}$$ converges. $$\begin{array}{c} 1 2 3 4 5 1 3 \end{array}$$ $$\begin{array}{ccc}{\displaystyle \sum _{n=1}^{\infty }\left|{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}\right|}& =& {\displaystyle \sum _{n=1}^{\infty }\frac{\text{log}\hspace{0.17em}n}{n}=\frac{\text{log}\hspace{0.17em}2}{2}+\frac{\text{log}\hspace{0.17em}3}{3}+\sum _{n=4}^{\infty }\frac{\text{log}\hspace{0.17em}n}{n}}\\ & <& {\displaystyle \frac{\text{log}\hspace{0.17em}2}{2}+\frac{\text{log}\hspace{0.17em}3}{3}+{\int }_4^{\infty }\frac{\text{log}\hspace{0.17em}n}{n}\text{diverges (from the previous page).}}\end{array}$$ So $\sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}$ is conditionally convergent but not absolutely convergent.

To apply the ratio test (to determine absolute convergence) one must analyse $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}\frac{\mid {\left(-1\right)}^{n+1}\frac{\text{log}\hspace{0.17em}n+1}{n+1}\mid }{\mid {\left(-1\right)}^n\frac{\text{log}\hspace{0.17em}n}{n}\mid }}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{n\text{log}(n+1)}{(n+1)\text{log}\hspace{0.17em}n}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{n(\text{log}(1+\frac{1}{n})-\text{log}\hspace{0.17em}n)}{(n+1)(\text{log}(1+\frac{1}{n-1})-\text{log}(n-1))}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{\left(\frac{n}{n}\right)\frac{\text{log}(1+\frac{1}{n})}{\frac{1}{n}}-n\text{log}\hspace{0.17em}n}{\left(\frac{n+1}{n-1}\right)\frac{\text{log}(1+\frac{1}{n-1})}{\frac{1}{n-1}}-(n+1)\text{log}(n-1)}}\end{array}$$ which is not very efficient. This makes sense since the series $\sum _{n=1}^{\infty }\frac{\text{log}\hspace{0.17em}n}{n}$ is not readily comparable to a series of the form $\sum _{n=1}^{\infty }{a}^n$ and ratio test is really a comparison to a series of this form.

Similarly the root test limit $$\underset{n\to \infty }{\text{lim}}{\left(\frac{\text{log}\hspace{0.17em}n}{n}\right)}^{\frac{1}{n}}=\underset{n\to \infty }{\text{lim}}\frac{{\left(\text{log}\hspace{0.17em}n\right)}^{\frac{1}{n}}}{n^{\frac{1}{n}}}=1$$ is not helpful for determining convergence.

Evaluate $\underset{n\to \infty }{\text{lim}}{n}^{\frac{1}{n}}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}{n}^{\frac{1}{n}}}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}{\left(e^{\text{log}\hspace{0.17em}n}\right)}^{\frac{1}{n}}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\frac{1}{n}\text{log}\hspace{0.17em}n}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\frac{1}{n}\text{log}(n-1+1)}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\frac{1}{n}(\text{log}(1+\frac{1}{n-1})+\text{log}(n-1))}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\frac{1}{n(n-1)}}\left(\frac{\text{log}(1+\frac{1}{n-1})}{\frac{1}{n-1}}\right)+\frac{1}{n}\text{log}(n-1)}\\ & =& {\displaystyle e^{0·1+0}=e^0=1\text{.}}\end{array}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100414Lect16.pdf and was given on 14 April 2010.

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