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We illustrate [link] in [link] . In particular, by representing the remainder ${R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\text{\cdots}$ 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,
and
We conclude that
Since
where ${S}_{N}$ is the $N\text{th}$ partial sum, we conclude that
Consider the series $\sum}_{n=1}^{\infty}1\text{/}{n}^{3}.$
For $\sum}_{n=1}^{\infty}\frac{1}{{n}^{4}},$ calculate ${S}_{5}$ and estimate the error ${R}_{5}.$
${S}_{5}\approx 1.09035,$ ${R}_{5}<0.00267$
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.
${a}_{n}=\frac{{2}^{n}+{3}^{n}}{{10}^{n\text{/}2}}$
${a}_{n}={e}^{\mathrm{-2}\text{/}n}$
$\underset{n\to \infty}{\text{lim}}{a}_{n}=1.$ 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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