MATH 409 Lecture 23

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Lecture Slides

Infinite Series

Given a sequence $\left\{a_n\right\}$ of real numbers, the expression

$a_1+a_2+\dots =\sum _{n=1}^{\mathrm{\infty }}$

Is called an infinite series with terms $a_n$

The partial sum of order $n$ is given by

$s_n=a_1+\dots +a_n$

If the sequence $\left\{s_n\right\}$ converges to limit $s\in \mathbb{ℝ}$, we say the series converges to $s$ or that $s$ is the sum of the series and write $\sum _{n=1}^{\mathrm{\infty }}{a}_n=s$

Otherwise the series diverges

Cauchy Criterion

Theorem. [Cauchy Criterion]. An infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges if and only if for every $ϵ>0$ there exists $N\in \mathbb{\mathbb{N}}$ such that $m\ge n\ge N$ implies $|a_n+a_{n+1}+\dots +a_m|<ϵ$

Proof. Let $\left\{s_n\right\}$ be the sequence of partial sums. Then

$a_n+a_n+1+\dots +a_m=s_m-s_{n-1}$

Consequently, the condition of the theorem is equivalent to the condition that $\left\{s_n\right\}$ be a Cauchy sequence. As we know, a sequence is convergent if and only if it is a Cauchy sequence.

quod erat demonstrandum

Examples

$\frac{1}{2}+\frac{1}{2^2}+\dots +\frac{1}{2^n}+\dots =1$

The partial sums $s_n$ of this series satisfy $s_n=1-2^{-n}$ for all $n\in \mathbb{\mathbb{N}}$. Thus $s_n\to 1$ as $n\to \mathrm{\infty }$

$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots +\frac{1}{n(n+1)}+\dots =1$

Since $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, the partial sums $s_n$ of this series satisfy $s_n=1-\frac{1}{n+1}$. Thus $s_n\to 1$ as $n\to \mathrm{\infty }$

$\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n=-1+1-1+\dots$ diverges

The partial sums $s_n$ satisfy $s_n=-1$ for odd $n$ and $s_n=0$ for even $n$. Hence the sequence $\left\{s_n\right\}$ has no limit.

The geometric series $\sum _{n=0}^{\mathrm{\infty }}{x}^n$ converges if and only if $\left|x\right|<1$, in which case its sum is $\frac{1}{1-x}$.

In the case $\left|x\right|\ge 1$, the series fails the divergence test. For any $x\ne 1$, the partial sums of the geometric series satisfy

$s_n=1+x+x^2+\dots +x^n=\frac{1-x^{n+1}}{1-x}$

For $\left|x\right|<1$, $s_n\to \frac{1}{1-x}$ as $n\to \mathrm{\infty }$.

Properties of Infinite Series

Theorem. [Divergence Test]. If the terms of an infinite series do not converge to zero, then the series diverges.

Theorem. [Linearity]. If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are convergent series, then

$\sum _{n=1}^{\mathrm{\infty }}(a_n+b_n)=\sum _{n=1}^{\mathrm{\infty }}{a}_n+\sum _{n=1}^{\mathrm{\infty }}{b}_n$

and

$\sum _{n=1}^{\mathrm{\infty }}\left(r{a}_n\right)=r\sum _{n=1}^{\mathrm{\infty }}{a}_n$

for any $r\in \mathbb{ℝ}$.

Theorem. If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ are convergent series, and $a_n\le b_n$ for all $n\in \mathbb{\mathbb{N}}$, then

$\sum _{n=1}^{\mathrm{\infty }}{a}_n\le \sum _{n=1}^{\mathrm{\infty }}{b}_n$

Series with Nonnegative Terms

Suppose that a series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ has nonnegative terms $a_n\ge 0$ for all $n\in \mathbb{\mathbb{N}}$. Then the sequence of partial sums $s_n=a_1+\dots +a_n$ is increasing. It follows that $\left\{s_n\right\}$

• converges to a finite limit if bounded and
• diverges to $+\mathrm{\infty }$ otherwise.

In the latter case, we write $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$.

Comparison Test

Theorem. [Comparison Test]. Suppose that $a_n,b_n\ge 0$ for all $n\in \mathbb{\mathbb{N}}$ and $a_n\le b_n$ for large enough $n$. Then

• convergence of the series $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ (larger terms) implies convergence of $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ (smaller terms), while
• divergence of the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$ (smaller terms) implies divergence of $\sum _{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$ (larger terms)

Proof. Since change a finite number of terms does not affect convergence of a series, it is no loss to assume that $a_n\le b_n$ for all $n\in \mathbb{\mathbb{N}}$. Then the partial sums $s_n=\sum _{k=1}^n{a}_k$ and $t_n=\sum _{k=1}^n{b}_n$ satisfy $s_n\le t_n$ for all $n$. Consequently, if $s_n\to +\mathrm{\infty }$ as $n\to \mathrm{\infty }$, then also $t_n\to +\mathrm{\infty }$ as $n\to \mathrm{\infty }$.

Conversely, if $\left\{t_n\right\}$ is bounded, then so is $\left\{s_n\right\}$.

Integral Test

Theorem. [Integral test]. Suppose a function $f:[1,\mathrm{\infty })\to \mathbb{ℝ}$ is positive and decreasing on $[1,\mathrm{\infty })$. Then

1. a sequence $\left\{y_n\right\}$ is bounded, where $y_n=f(1)+f(2)+\dots +f(n)-\underset{1}{\overset{n}{\int }}f(x)\mathrm{d}x$ $n=1,2,\dots$
2. the series $\sum _{n=1}^{\mathrm{\infty }}f(n)$ is convergent if and only if the function $f$ is improperly integrable on $[1,\mathrm{\infty })$.

Proof. To prove the theorem, we need the following lemma:

Lemma. Any monotone function $g:[a,b]\to \mathbb{ℝ}$ is integrable on $[a,b]$.

Idea of the proof. Any monotone function has only jump discontinuities. Further, any function has at most countably many jump discontinuities. Besides, a monotone function on a closed interval $[a,b]$ is clearly bounded.

The lemma implies that the function $f$ is integrable on every closed interval $J=[a,b]\subset [1,\mathrm{\infty })$. Then for any partition $P$ of the interval $J$, the lower Darboux sum $L(f,P)$ and the upper Darboux sum $U(f,P)$ satisfy

$L(f,P)\le \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le U(f,P)$

Let $P=\{x_0,x_1,\dots ,x_k\}$, where $x_0<x_1<\dots <x_k$. Then $\sup f\left([x_{j-1},x_j]\right)=f(x_{j-1})$ and $\inf f\left([x_{j-1},x_j]\right)=f(x_j)$ since $f$ is decreasing. In the case $J=[1,n]$, where $n\in \mathbb{\mathbb{N}}$, and $P=\{1,2,\dots ,n\}$, we obtain

• $L(f,P)=f(2)+f(3)+\dots +f(n)$,
• $U(f,P)=f(1)+f(2)+\dots +f(n-1)$.

Then the above inequalities imply that $0<f(n)\le y_n\le f(1)$. Thus the sequence $\left\{y_n\right\}$ is bounded.

Now for the second part of the theorem: Since $f$ is positive, the series $\sum _{n=1}^{\mathrm{\infty }}f(n)$ either converges or else it diverges to $+\mathrm{\infty }$. Likewise, the improper integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)\mathrm{d}x$ either converges or else it diverges to $+\mathrm{\infty }$. Since the sequence $\left\{y_n\right\}$ is bounded by the above, divergence of the series and the integral imply each other.

quod erat demonstrandum

Examples

Theorem. [P-Series Test]. [Riemann Zeta Function]. $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$ is convergent for any $p>1$ and divergent for $p<1$.

Proof. For any $p\ne 1$, we have $\int x^{-p}\mathrm{d}x=\frac{x^{1-p}}{1-p}+C$ on the interval $[1,\mathrm{\infty })$. The antiderivative converges to a finite limit at $+\mathrm{\infty }$ in the case $p>1$ and diverges to $+\mathrm{\infty }$ for $p<1$. Hence the function $f(x)=x^{-p}$ is improperly integrable on $[1,\mathrm{\infty })$ for $p>1$, but not for $p<1$. By the Integral Test, the series is convergent for $p>1$ and divergent for $0\le p<1$.

If $p<0$ then the Integral Test does not apply since $f$ is not decreasing. In this case, the series is divergent since the terms $\frac{1}{n^p}$ do not converge to $0$ as $n\to \mathrm{\infty }$.

quod erat demonstrandum

Harmonic Series. $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ diverges.

Indeed $\underset{1}{\overset{n}{\int }}\frac{1}{x}\mathrm{d}x=\log x\to +\mathrm{\infty }$ as $n\to 0$. By the integral test, the series is divergent.

Moreover, the sequence $y_n=\sum _{k=1}^n{k}^{-1}-\log n$ is bounded (actually, it is decreasing and hence convergent)

This was the bonus problem on the test.

$\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n{\log }^{2}n}$ converges.

The antiderivative of $f(x)={\left(x{\log }^{2}x\right)}^{-1}$ on $(1,\mathrm{\infty })$ is

$\int \frac{1}{x{\log }^{2}x}\mathrm{d}x=-\frac{1}{\log x}+C$

Since the antiderivative converges to a finite limit at $+\mathrm{\infty }$, the function $f$ is improperly integrable on $[2,\mathrm{\infty })$.

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{1+n^2}$ converges.

Indeed, $0<\frac{1}{1+n^2}\le \frac{1}{n^2}$ for all $n\in \mathbb{\mathbb{N}}$. Since the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$ is convergent, it remains to apply the comparison test. Alternatively, we can use the integral test.

$\int \frac{1}{1+x^2}\mathrm{d}x=\arctan x+C$ converges to a finite limit ($\frac{\pi }{2}+C$) at $+\mathrm{\infty }$ so the function $f(x)=\frac{1}{1+x^2}$ is improperly integrable on $[1,\mathrm{\infty })$.

$\sum _{n=1}^{\mathrm{\infty }}{\mathrm{e}}^{-n^2}$ converges (really fast!)

We have $0<{\mathrm{e}}^{-n^2}\le {\mathrm{e}}^{-n}$ for all $n\in \mathbb{\mathbb{N}}$. The geometric series $\sum _{n=1}^{\mathrm{\infty }}{\mathrm{e}}^{-n}=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{e}\right)}^n$ is convergent since $0<{\mathrm{e}}^{-1}<1$. By the comparison test, $\sum _{n=1}^{\mathrm{\infty }}{\mathrm{e}}^{-n^2}$ is convergent as well.