# MATH 409 Lecture 24

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Lecture Slides

## 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 ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges if and only if for every ${\displaystyle \epsilon >0}$, there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle m\geq n\geq N}$ implies ${\displaystyle \left|a_{n}+a_{n+1}+\dots +a_{m}\right|<\epsilon }$
comparison test (series with nonnegative terms)
Suppose ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle a_{n} for large ${\displaystyle n}$. Then convergence of the series ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$ implies convergence of ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ while ${\displaystyle \sum _{n=1}^{\infty }a_{n}=\infty }$ implies ${\displaystyle \sum _{n=1}^{\infty }b_{n}=\infty }$
integral test
Suppose that a function ${\displaystyle f:\left[1,\infty \right)\to \mathbb {R} }$ is positive and decreasing on ${\displaystyle \left[1,\infty \right)}$ then the series ${\displaystyle \sum _{n=1}^{\infty }f(n)}$ converges if and only if the function ${\displaystyle f}$ is improperly integrable on ${\displaystyle \left[1,\infty \right)}$

## Alternating Series

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

### Leibniz Criterion

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

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

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

${\displaystyle s_{2n}=s_{2n-1}-a_{2n}

Since the sequence ${\displaystyle \left\{a_{n}\right\}}$ is decreasing, we also have

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

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

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

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

This series converges due to the alternating series test.

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

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

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

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

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

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

## Absolute Convergence of Series

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

Theorem. Any absolutely convergent series is convergent.

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

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

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

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

quod erat demonstrandum

### Examples

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

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

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

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

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

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

## Other Tests

### Ratio Test (d'Alembert's Criterion)

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

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

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

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

quod erat demonstrandum

### Root Test

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

${\displaystyle r=\limsup _{n\to \infty }{\sqrt[{n}]{\left|a_{n}\right|}}}$
1. If ${\displaystyle r<1}$, then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges absolutely
2. If ${\displaystyle r>1}$, then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ diverges.

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

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

quod erat demonstrandum

### Examples

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

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

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

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

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

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

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

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