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 $\sum _{n=1}^{\infty }a_{n}$ converges if and only if for every $\epsilon >0$, there exists $N\in \mathbb {N}$ such that $m\geq n\geq N$ implies $\left|a_{n}+a_{n+1}+\dots +a_{m}\right|<\epsilon$

comparison test (series with nonnegative terms)

Suppose $a_{n},b_{n}\geq 0$ for all $n\in \mathbb {N}$ and $a_{n}<b_{n}$ for large $n$. Then convergence of the series $\sum _{n=1}^{\infty }b_{n}$ implies convergence of $\sum _{n=1}^{\infty }a_{n}$ while $\sum _{n=1}^{\infty }a_{n}=\infty$ implies $\sum _{n=1}^{\infty }b_{n}=\infty$

integral test

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

Alternating Series

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

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 $n\to \infty$, then the following alternating series converges:

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

Proof. Let $s_{n}=\sum _{k=1}^{n}\left(-1\right)^{k+1}\,a_{k}$ the the partial sum of order $n$ of the series. For any $n\in \mathbb {N}$ 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

${\begin{aligned}s_{2n+1}&=s_{2n-1}-a_{2n}+a_{2n+1}\leq s_{2n-1}\\s_{2n+2}&=s_{2n}+a_{2n+1}-a_{2n+2}\geq s_{2n}\end{aligned}}$

Therefore $s_{2n}\leq s_{2n+2}<s_{2n+1}\leq s_{2n-1}$ for all $n\in \mathbb {N}$. 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 $n\to \infty$, 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

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

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]. $\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n}}{2n-1}}=1+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{7}}-\dots$

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)

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

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 $\sum _{n=1}^{\infty }a_{n}$ is said to converge absolutely if $\sum _{n=1}^{\infty }\left|a_{n}\right|<\infty$.

Theorem. Any absolutely convergent series is convergent.

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

$\left|\left|a_{n}\right|+\left|a_{n+1}\right|+\dots +\left|a_{m}\right|\right|<\epsilon$

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

Then $\left|a_{n}+a_{n+1}+\dots +a_{m}\right|\leq \left|a_{n}\right|+\left|a_{n+1}\right|+\dots +\left|a_{m}\right|<\epsilon$ (by the triangle inequality) for $m\geq n\geq N$. According to the Cauchy criterion, the series $\sum _{n=1}^{\infty }a_{n}$ converges.

quod erat demonstrandum

Examples

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

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

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

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

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

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

Other Tests

Ratio Test (d'Alembert's Criterion)

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

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

1. If $r<1$, then $\sum _{n=1}^{\infty }a_{n}$ converges absolutely
2. If $r>1$, then $\sum _{n=1}^{\infty }a_{n}$ diverges.
3. If $r=1$, the test is inconclusive. (e.g. $\lim _{n\to \infty }{\frac {(n+1)^{-p}}{n^{-p}}}=1$, 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 $n\to \infty$ so that the series $\sum _{n=1}^{\infty }a_{n}$ diverges due to the divergence test.

In the case $r<1$, choose some $x\in \left(r,1\right)$. 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 $n\in \mathbb {N}$. Since $0<r<x<1$, the geometric series $\sum _{n=1}^{\infty }x^{n}$ converges. So does the series $\sum _{n=1}^{\infty }C\,x^{n}$. By the comparison test, the series $\sum _{n=1}^{\infty }\left|a_{n}\right|$ converges as well. The convergence of $\sum _{n=1}^{\infty }a_{n}$ follows from absolute convergence.

quod erat demonstrandum

Root Test

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

$r=\limsup _{n\to \infty }{\sqrt[{n}]{\left|a_{n}\right|}}$

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

Proof. If $r>1$, then $\sup _{k\geq n}{\sqrt[{k}]{\left|a_{k}\right|}}\geq r>1$ for all $n\in \mathbb {N}$. Therefore for any $n\in \mathbb {N}$, there exists $k(n)\geq n$ 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 (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_k \nto 0} as $k\to \infty$ so that the series $\sum _{k=1}^{\infty }a_{k}$ diverges due to the divergence test.

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

quod erat demonstrandum

Examples

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

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

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

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

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

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

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

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