MATH 409 Lecture 24

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Lecture Slides

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Tests of Convergence

divergence test

If the terms of an infinite series do not converge to zero, then the series diverges

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 $|a_n+a_{n+1}+\dots +a_m|<ϵ$

comparison test (series with nonnegative terms)

Suppose $a_n,b_n\ge 0$ for all $n\in \mathbb{\mathbb{N}}$ and $a_n<b_n$ for large $n$. Then convergence of the series $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ implies convergence of $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ while $\sum _{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$ implies $\sum _{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$

integral test

Suppose that a function $f:[1,\mathrm{\infty })\to \mathbb{ℝ}$ is positive and decreasing on $[1,\mathrm{\infty })$ then the series $\sum _{n=1}^{\mathrm{\infty }}f(n)$ converges if and only if the function $f$ is improperly integrable on $[1,\mathrm{\infty })$

Alternating Series

An infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is called alternating if any two neighboring terms have different signs: $a_n{a}_{n+1}<0$ for all $n\in \mathbb{\mathbb{N}}$.

Leibniz Criterion

Theorem. [Alternating Series Test]. [Leibniz Criterion]. If $\left\{a_n\right\}$ is a decreasing sequence of positive numbers and $a_n\to 0$ as $n\to \mathrm{\infty }$, then the following alternating series converges:

$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n=a_1-a_2+a_3-\dots$

Proof. Let $s_n=\sum _{k=1}^n{(-1)}^{k+1}{a}_k$ the the partial sum of order $n$ of the series. For any $n\in \mathbb{\mathbb{N}}$ we have (odd terms are positive, even terms are negative)

$s_{2n}=s_{2n-1}-a_{2n}<s_{2n-1}$

Since the sequence $\left\{a_n\right\}$ is decreasing, we also have

$\begin{array}{rl}{s}_{2n+1}& =s_{2n-1}-a_{2n}+a_{2n+1}\le s_{2n-1}\\ s_{2n+2}& =s_{2n}+a_{2n+1}-a_{2n+2}\ge s_{2n}\end{array}$

Therefore $s_{2n}\le s_{2n+2}<s_{2n+1}\le s_{2n-1}$ for all $n\in \mathbb{\mathbb{N}}$. It follows that a subsequence $\left\{s_{2n}\right\}$ is increasing, and a subsequence $\left\{s_{2n-1}\right\}$ is decreasing, and both are bounded. Hence both subsequences are convergent. (but do they both converge to the same limit?)

Since $s_{2n-1}-s_{2n}=a_{2n}\to 0$ as $n\to \mathrm{\infty }$, both subsequences converge to the same limit $L$, then $L$ is the limit of the entire sequence $\left\{s_n\right\}$.

quod erat demonstrandum

Note: This theorem is not directly proved in the book. Rather it is a corollary of a more general theorem, Abel's Theorem, which is more complicated.

Examples

$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\dots$

This series converges due to the alternating series test.

One can show that the sum is $\log 2$ (this is a functional series in the form of a taylor series).

[Leibniz series]. $\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n-1}=1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\dots$

After multiplying all terms by $-1$, the series satisfies all conditions of the alternating series test.

It follows that the series converges to $-\frac{\pi }{4}$ (this is also a functional series in the form of a fourier series)

$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}n}{2n-1}=1-\frac{2}{3}+\frac{3}{5}-\frac{4}{7}+\dots$

This series diverges because even though the terms decrease in absolute value, they converge to $\frac{1}{2}$, not $0$

Absolute Convergence of Series

An infinite series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is said to converge absolutely if $\sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|<\mathrm{\infty }$.

Theorem. Any absolutely convergent series is convergent.

Proof. Suppose that a series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely, that is, the series $\sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|$ converges. By the Cauchy criterion, for every $ϵ>0$, there exists $N\in \mathbb{\mathbb{N}}$ such that

$|\left|a_n\right|+\left|a_{n+1}\right|+\dots +\left|a_m\right||<ϵ$

for $m\ge n\ge N$. (The outer absolute values are redundant since all terms inside will be positive anyway)

Then $|a_n+a_{n+1}+\dots +a_m|\le \left|a_n\right|+\left|a_{n+1}\right|+\dots +\left|a_m\right|<ϵ$ (by the triangle inequality) for $m\ge n\ge N$. According to the Cauchy criterion, the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges.

quod erat demonstrandum

Examples

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}=1+\frac{1}{2^3}+\frac{1}{3^3}+\dots$

The series converges due to the integral test. Since it has positive terms, it is absolutely convergent as well.

$\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n}{n^2}=\sin 1+\frac{\sin 2}{4}+\frac{\sin 3}{4}+\dots$

This series converges since $\left|\frac{\sin n}{n^2}\right|\le \frac{1}{n^2}$ converges absolutely by the comparison test.

$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\dots$

This series converges due to the alternating series test, but it is not absolutely convergent as the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ diverges.

Other Tests

Ratio Test (d'Alembert's Criterion)

Theorem. Let $\left\{a_n\right\}$ be a sequence of real numbers with $a_n\ne 0$ for large $n$. Suppose that the following limit exists (finite or infinite):

$r=\underset{n\to \mathrm{\infty }}{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}$

1. If $r<1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely
2. If $r>1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.
3. If $r=1$, the test is inconclusive. (e.g. $\underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1{)}^{-p}}{n^{-p}}=1$, but $n^{-p}$ converges for $p>1$ and diverges otherwise.)

Proof. If $r>1$, then $\frac{\left|a_{n+1}\right|}{\left|a_n\right|}>1$ for $n$ large enough. It follows that the sequence $\left\{\left|a_n\right|\right\}$ is eventually increasing. Then Failed to parse (unknown function "\nto"): {\displaystyle a_n \nto 0} as $n\to \mathrm{\infty }$ so that the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges due to the divergence test.

In the case $r<1$, choose some $x\in (r,1)$. Then $\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<x$ for $n$ large enough. Consequently, $\frac{\left|a_{n+1}\right|}{x^{n+1}}<\frac{\left|a_n\right|}{x^n}$ for $n$ large enough. That is, the sequence $\left\{\frac{\left|a_n\right|}{x^n}\right\}$ is eventually decreasing. It follows that this sequence is bounded. Hence $\left|a_n\right|<C{x}^n$ for some $C>0$ and all $n\in \mathbb{\mathbb{N}}$. Since $0<r<x<1$, the geometric series $\sum _{n=1}^{\mathrm{\infty }}{x}^n$ converges. So does the series $\sum _{n=1}^{\mathrm{\infty }}C{x}^n$. By the comparison test, the series $\sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|$ converges as well. The convergence of $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ follows from absolute convergence.

quod erat demonstrandum

Root Test

Theorem. Let $\left\{a_n\right\}$ be a sequence of real numbers and

$r=\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\sqrt[n]{\left|a_n\right|}$

1. If $r<1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely
2. If $r>1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.

Proof. If $r>1$, then $\underset{k\ge n}{\sup }\sqrt[k]{\left|a_k\right|}\ge r>1$ for all $n\in \mathbb{\mathbb{N}}$. Therefore for any $n\in \mathbb{\mathbb{N}}$, there exists $k(n)\ge n$ such that ${\left|a_{k(n)}\right|}^{\frac{1}{k(n)}}>1$. In particular, $\left|a_{k(n)}\right|>1$. It follows that Failed to parse (unknown function "\nto"): {\displaystyle a_k \nto 0} as $k\to \mathrm{\infty }$ so that the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ diverges due to the divergence test.

In the case $r<1$, choose some $x\in (r,1)$. Then $\underset{k\ge n}{\sup }\sqrt[k]{\left|a_k\right|}<x$ for some $n\in \mathbb{\mathbb{N}}$. This implies that $\left|a_k\right|<x^k$ for all $k\ge n$. Since $0<r<x<1$, the geometric series $\sum _{k=1}^{\mathrm{\infty }}{x}^k$ converges. By the comparison test, the series $\sum _{k=1}^{\mathrm{\infty }}\left|a_k\right|$ converges as well. The convergence of $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ follows from absolute convergence.

quod erat demonstrandum

Examples

$\sum _{n=1}^{\mathrm{\infty }}\frac{n}{2^n}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots$

If $a_n=\frac{n}{2^n}$, then $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \mathrm{\infty }}{\lim }\frac{n+1}{2n}=\frac{1}{2}$. The series converges by the ratio test.

$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$, where $x\in \mathbb{ℝ}$

In the case $x=0$, we have a finite sum. In the case $x\ne 0$, let $a_n=\frac{x^n}{n!}$, then $\underset{n\to \mathrm{\infty }}{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\left|x\right|}{n+1}=0$. Therefore the series converges absolutely for all $x\ne 0$ by the ratio test.

$\sum _{n=1}^{\mathrm{\infty }}\frac{{(n!)}^2}{\left(2n\right)!}=\frac{{(1!)}^2}{2!}+\frac{{(2!)}^2}{4!}+\frac{(3!{)}^2}{6!}+\dots$

If $a_n=\frac{{(n!)}^2}{\left(2n\right)!}$, then $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_{n+1}}{a_n}=\frac{1+n^{-1}}{4+2n^{-1}}=\frac{1}{4}$. By the ratio test, the series converges.

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{n}{n+1}\right)}^{n^2}=\frac{1}{2}+{\left(\frac{2}{3}\right)}^4+{\left(\frac{3}{4}\right)}^9+\dots$

If $a_n={\left(\frac{n}{n+1}\right)}^{n^2}$, then $\underset{n\to \mathrm{\infty }}{\lim }\sqrt[n]{a_n}=\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{n}{n+1}\right)}^n=\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^{-n}=\frac{1}{\mathrm{e}}$. By the root test, the series converges.