MATH 409 Lecture 23

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Lecture Slides

Infinite Series

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

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

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 $\sum _{n=1}^{\infty }a_{n}=s$

Otherwise the series diverges

Cauchy Criterion

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

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 $n\in \mathbb {N}$. Thus $s_{n}\to 1$ as $n\to \infty$

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

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 $n\to \infty$

$\sum _{n=1}^{\infty }\left(-1\right)^{n}=-1+1-1+\dots$ 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 $\sum _{n=0}^{\infty }x^{n}$ converges if and only if $\left|x\right|<1$, in which case its sum is ${\frac {1}{1-x}}$.

In the case $\left|x\right|\geq 1$, the series fails the divergence test. For any $x\neq 1$, 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 $n\to \infty$.

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 $\sum _{n=1}^{\infty }a_{n}$ and $\sum _{n=1}^{\infty }b_{n}$ are convergent series, then

$\sum _{n=1}^{\infty }(a_{n}+b_{n})=\sum _{n=1}^{\infty }a_{n}+\sum _{n=1}^{\infty }b_{n}$

and

$\sum _{n=1}^{\infty }\left(r\,a_{n}\right)=r\,\sum _{n=1}^{\infty }a_{n}$

for any $r\in \mathbb {R}$.

Theorem. If $\sum _{n=1}^{\infty }a_{n}$ and $\sum _{n=1}^{\infty }b_{n}$ are convergent series, and $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$, then

$\sum _{n=1}^{\infty }a_{n}\leq \sum _{n=1}^{\infty }b_{n}$

Series with Nonnegative Terms

Suppose that a series $\sum _{n=1}^{\infty }a_{n}$ has nonnegative terms $a_{n}\geq 0$ for all $n\in \mathbb {N}$. 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 $+\infty$ otherwise.

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

Comparison Test

Theorem. [Comparison Test]. Suppose that $a_{n},b_{n}\geq 0$ for all $n\in \mathbb {N}$ and $a_{n}\leq b_{n}$ for large enough $n$. Then

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

Proof. Since change a finite number of terms does not affect convergence of a series, it is no loss to assume that $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$. Then the partial sums $s_{n}=\sum _{k=1}^{n}a_{k}$ and $t_{n}=\sum _{k=1}^{n}b_{n}$ satisfy $s_{n}\leq t_{n}$ for all $n$. Consequently, if $s_{n}\to +\infty$ as $n\to \infty$, then also $t_{n}\to +\infty$ as $n\to \infty$.

Conversely, if $\left\{t_{n}\right\}$ is bounded, then so is $\left\{s_{n}\right\}$.

Integral Test

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

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

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

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

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 $\left[a,b\right]$ is clearly bounded.

The lemma implies that the function $f$ is integrable on every closed interval $J=[a,b]\subset \left[1,\infty \right)$. 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

$L(f,P)\leq \int _{a}^{b}f(x)\,\mathrm {d} x\leq U(f,P)$

Let $P=\left\{x_{0},x_{1},\ldots ,x_{k}\right\}$, where $x_{0}<x_{1}<\dots <x_{k}$. Then $\sup {f\left(\left[x_{j-1},x_{j}\right]\right)}=f(x_{j-1})$ and $\inf {f\left(\left[x_{j-1},x_{j}\right]\right)}=f(x_{j})$ since $f$ is decreasing. In the case $J=\left[1,n\right]$, where $n\in \mathbb {N}$, and $P=\left\{1,2,\ldots ,n\right\}$, 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 $0<f(n)\leq y_{n}\leq f(1)$. Thus the sequence $\left\{y_{n}\right\}$ is bounded.

Now for the second part of the theorem: Since $f$ is positive, the series $\sum _{n=1}^{\infty }f(n)$ either converges or else it diverges to $+\infty$. Likewise, the improper integral $\int _{1}^{\infty }f(x)\,\mathrm {d} x$ either converges or else it diverges to $+\infty$. 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]. $\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}$ is convergent for any $p>1$ and divergent for $p<1$.

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

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 $n\to \infty$.

quod erat demonstrandum

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

Indeed $\int _{1}^{n}{\frac {1}{x}}\,\mathrm {d} x=\log {x}\to +\infty$ 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.

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

The antiderivative of $f(x)=\left(x\,\log ^{2}{x}\right)^{-1}$ on $(1,\infty )$ 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 $+\infty$, the function $f$ is improperly integrable on $\left[2,\infty \right)$.

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

Indeed, $0<{\frac {1}{1+n^{2}}}\leq {\frac {1}{n^{2}}}$ for all $n\in \mathbb {N}$. Since the series $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ 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 $+\infty$ so the function $f(x)={\frac {1}{1+x^{2}}}$ is improperly integrable on $\left[1,\infty \right)$.

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

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