# 5.3 The divergence and integral tests  (Page 4/9)

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We illustrate [link] in [link] . In particular, by representing the remainder ${R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{⋯}$ as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by ${\int }_{N}^{\infty }f\left(x\right)dx$ and bounded below by ${\int }_{N+1}^{\infty }f\left(x\right)dx.$ In other words,

${R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{⋯}>{\int }_{N+1}^{\infty }f\left(x\right)dx$

and

${R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{⋯}<{\int }_{N}^{\infty }f\left(x\right)dx.$

We conclude that

${\int }_{N+1}^{\infty }f\left(x\right)dx<{R}_{N}<{\int }_{N}^{\infty }f\left(x\right)dx.$

Since

$\sum _{n=1}^{\infty }{a}_{n}={S}_{N}+{R}_{N},$

where ${S}_{N}$ is the $N\text{th}$ partial sum, we conclude that

${S}_{N}+{\int }_{N+1}^{\infty }f\left(x\right)dx<\sum _{n=1}^{\infty }{a}_{n}<{S}_{N}+{\int }_{N}^{\infty }f\left(x\right)dx.$

## Estimating the value of a series

Consider the series $\sum _{n=1}^{\infty }1\text{/}{n}^{3}.$

1. Calculate ${S}_{10}=\sum _{n=1}^{10}1\text{/}{n}^{3}$ and estimate the error.
2. Determine the least value of $N$ necessary such that ${S}_{N}$ will estimate $\sum _{n=1}^{\infty }1\text{/}{n}^{3}$ to within $0.001.$
1. Using a calculating utility, we have
${S}_{10}=1+\frac{1}{{2}^{3}}+\frac{1}{{3}^{3}}+\frac{1}{{4}^{3}}+\text{⋯}+\frac{1}{{10}^{3}}\approx 1.19753.$

By the remainder estimate, we know
${R}_{N}<{\int }_{N}^{\infty }\frac{1}{{x}^{3}}dx.$

We have
${\int }_{10}^{\infty }\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}{\int }_{10}^{b}\frac{1}{{x}^{3}}dx=\underset{b\to \infty }{\text{lim}}\left[-\frac{1}{2{x}^{2}}{\right]}_{N}^{b}=\underset{b\to \infty }{\text{lim}}\left[-\frac{1}{2{b}^{2}}+\frac{1}{2{N}^{2}}\right]=\frac{1}{2{N}^{2}}.$

Therefore, the error is ${R}_{10}<1\text{/}2{\left(10\right)}^{2}=0.005.$
2. Find $N$ such that ${R}_{N}<0.001.$ In part a. we showed that ${R}_{N}<1\text{/}2{N}^{2}.$ Therefore, the remainder ${R}_{N}<0.001$ as long as $1\text{/}2{N}^{2}<0.001.$ That is, we need $2{N}^{2}>1000.$ Solving this inequality for $N,$ we see that we need $N>22.36.$ To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is $N=23.$

For $\sum _{n=1}^{\infty }\frac{1}{{n}^{4}},$ calculate ${S}_{5}$ and estimate the error ${R}_{5}.$

${S}_{5}\approx 1.09035,$ ${R}_{5}<0.00267$

## Key concepts

• If $\underset{n\to \infty }{\text{lim}}{a}_{n}\ne 0,$ then the series $\sum _{n=1}^{\infty }{a}_{n}$ diverges.
• If $\underset{n\to \infty }{\text{lim}}{a}_{n}=0,$ the series $\sum _{n=1}^{\infty }{a}_{n}$ may converge or diverge.
• If $\sum _{n=1}^{\infty }{a}_{n}$ is a series with positive terms ${a}_{n}$ and $f$ is a continuous, decreasing function such that $f\left(n\right)={a}_{n}$ for all positive integers $n,$ then
$\sum _{n=1}^{\infty }{a}_{n}\text{and}{\int }_{1}^{\infty }f\left(x\right)dx$

either both converge or both diverge. Furthermore, if $\sum _{n=1}^{\infty }{a}_{n}$ converges, then the $N\text{th}$ partial sum approximation ${S}_{N}$ is accurate up to an error ${R}_{N}$ where ${\int }_{N+1}^{\infty }f\left(x\right)dx<{R}_{N}<{\int }_{N}^{\infty }f\left(x\right)dx.$
• The p -series $\sum _{n=1}^{\infty }1\text{/}{n}^{p}$ converges if $p>1$ and diverges if $p\le 1.$

## Key equations

• Divergence test
$\text{If}\phantom{\rule{0.2em}{0ex}}{a}_{n}↛0\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}n\to \infty ,\sum _{n=1}^{\infty }{a}_{n}\phantom{\rule{0.2em}{0ex}}\text{diverges}.$
• p -series
$\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}\left\{\begin{array}{l}\text{converges if}\phantom{\rule{0.2em}{0ex}}p>1\\ \text{diverges if}\phantom{\rule{0.2em}{0ex}}p\le 1\end{array}$
• Remainder estimate from the integral test
${\int }_{N+1}^{\infty }f\left(x\right)dx<{R}_{N}<{\int }_{N}^{\infty }f\left(x\right)dx$

For each of the following sequences, if the divergence test applies, either state that $\underset{n\to \infty }{\text{lim}}{a}_{n}$ does not exist or find $\underset{n\to \infty }{\text{lim}}{a}_{n}.$ If the divergence test does not apply, state why.

${a}_{n}=\frac{n}{n+2}$

${a}_{n}=\frac{n}{5{n}^{2}-3}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=0.$ Divergence test does not apply.

${a}_{n}=\frac{n}{\sqrt{3{n}^{2}+2n+1}}$

${a}_{n}=\frac{\left(2n+1\right)\left(n-1\right)}{{\left(n+1\right)}^{2}}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=2.$ Series diverges.

${a}_{n}=\frac{{\left(2n+1\right)}^{2n}}{{\left(3{n}^{2}+1\right)}^{n}}$

${a}_{n}=\frac{{2}^{n}}{{3}^{n\text{/}2}}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=\infty$ (does not exist). Series diverges.

${a}_{n}=\frac{{2}^{n}+{3}^{n}}{{10}^{n\text{/}2}}$

${a}_{n}={e}^{-2\text{/}n}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=1.$ Series diverges.

${a}_{n}=\text{cos}\phantom{\rule{0.1em}{0ex}}n$

${a}_{n}=\text{tan}\phantom{\rule{0.1em}{0ex}}n$

$\underset{n\to \infty }{\text{lim}}{a}_{n}$ does not exist. Series diverges.

${a}_{n}=\frac{1-{\text{cos}}^{2}\left(1\text{/}n\right)}{{\text{sin}}^{2}\left(2\text{/}n\right)}$

${a}_{n}={\left(1-\frac{1}{n}\right)}^{2n}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=1\text{/}{e}^{2}.$ Series diverges.

${a}_{n}=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{n}$

${a}_{n}=\frac{{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{2}}{\sqrt{n}}$

$\underset{n\to \infty }{\text{lim}}{a}_{n}=0.$ Divergence test does not apply.

State whether the given $p$ -series converges.

$\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$

$\sum _{n=1}^{\infty }\frac{1}{n\sqrt{n}}$

Series converges, $p>1.$

$\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{2}}}$

$\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{{n}^{4}}}$

Series converges, $p=4\text{/}3>1.$

$\sum _{n=1}^{\infty }\frac{{n}^{e}}{{n}^{\pi }}$

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