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$\sum}_{n=1}^{\infty}\frac{1}{2\left(n+1\right)$
$\sum}_{n=1}^{\infty}\frac{1}{2n-1$
Diverges by comparison with harmonic series, since $2n-1\ge n.$
$\sum}_{n=2}^{\infty}\frac{1}{{\left(n\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{2}$
$\sum}_{n=1}^{\infty}\frac{n\text{!}}{\left(n+2\right)\text{!}$
${a}_{n}=1\text{/}\left(n+1\right)\left(n+2\right)<1\text{/}{n}^{2}.$ Converges by comparison with p -series, $p=2.$
$\sum}_{n=1}^{\infty}\frac{1}{n\text{!}$
$\sum _{n=1}^{\infty}\frac{\text{sin}\left(1\text{/}n\right)}{n}$
$\text{sin}\left(1\text{/}n\right)\le 1\text{/}n,$ so converges by comparison with p -series, $p=2.$
$\sum _{n=1}^{\infty}\frac{{\text{sin}}^{2}n}{{n}^{2}}$
$\sum _{n=1}^{\infty}\frac{\text{sin}\left(1\text{/}n\right)}{\sqrt{n}}$
$\text{sin}\left(1\text{/}n\right)\le 1,$ so converges by comparison with p -series, $p=3\text{/}2.$
$\sum}_{n=1}^{\infty}\frac{{n}^{1.2}-1}{{n}^{2.3}+1$
$\sum}_{n=1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n$
Since $\sqrt{n+1}-\sqrt{n}=1\text{/}\left(\sqrt{n+1}+\sqrt{n}\right)\le 2\text{/}\sqrt{n},$ series converges by comparison with p -series for $p=\mathrm{1.5.}$
$\sum}_{n=1}^{\infty}\frac{\sqrt[4]{n}}{\sqrt[3]{{n}^{4}+{n}^{2}}$
Use the limit comparison test to determine whether each of the following series converges or diverges.
$\sum}_{n=1}^{\infty}{\left(\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{n}\right)}^{2$
Converges by limit comparison with p -series for $p>1.$
$\sum}_{n=1}^{\infty}{\left(\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}n}{{n}^{0.6}}\right)}^{2$
$\sum}_{n=1}^{\infty}\frac{\text{ln}\left(1+\frac{1}{n}\right)}{n$
Converges by limit comparison with p -series, $p=2.$
$\sum}_{n=1}^{\infty}\text{ln}\left(1+\frac{1}{{n}^{2}}\right)$
$\sum}_{n=1}^{\infty}\frac{1}{{4}^{n}-{3}^{n}$
Converges by limit comparison with ${4}^{\text{\u2212}n}.$
$\sum}_{n=1}^{\infty}\frac{1}{{n}^{2}-n\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}n$
$\sum}_{n=1}^{\infty}\frac{1}{{e}^{\left(1.1\right)n}-{3}^{n}$
Converges by limit comparison with $1\text{/}{e}^{1.1n}.$
$\sum}_{n=1}^{\infty}\frac{1}{{e}^{\left(1.01\right)n}-{3}^{n}$
$\sum}_{n=1}^{\infty}\frac{1}{{n}^{1+1\text{/}n}$
Diverges by limit comparison with harmonic series.
$\sum}_{n=1}^{\infty}\frac{1}{{2}^{1+1\text{/}n}{n}^{1+1\text{/}n}$
$\sum}_{n=1}^{\infty}\left(\frac{1}{n}-\text{sin}\left(\frac{1}{n}\right)\right)$
Converges by limit comparison with p -series, $p=3.$
$\sum}_{n=1}^{\infty}\left(1-\text{cos}\left(\frac{1}{n}\right)\right)$
$\sum}_{n=1}^{\infty}\frac{1}{n}\left({\text{tan}}^{\mathrm{-1}}n-\frac{\pi}{2}\right)$
Converges by limit comparison with p -series, $p=3.$
$\sum}_{n=1}^{\infty}{\left(1-\frac{1}{n}\right)}^{n.n$ ( Hint: ${\left(1-\frac{1}{n}\right)}^{n}\to 1\text{/}e.)$
$\sum}_{n=1}^{\infty}\left(1-{e}^{\mathrm{-1}\text{/}n}\right)$ ( Hint: $1\text{/}e\approx {\left(1-1\text{/}n\right)}^{n},$ so $1-{e}^{\mathrm{-1}\text{/}n}\approx 1\text{/}n.)$
Diverges by limit comparison with $1\text{/}n.$
Does $\sum}_{n=2}^{\infty}\frac{1}{{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{p}$ converge if $p$ is large enough? If so, for which $p\text{?}$
Does $\sum}_{n=1}^{\infty}{\left(\frac{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}{n}\right)}^{p$ converge if $p$ is large enough? If so, for which $p\text{?}$
Converges for $p>1$ by comparison with a $p$ series for slightly smaller $p.$
For which $p$ does the series $\sum}_{n=1}^{\infty}{2}^{pn}\text{/}{3}^{n$ converge?
For which $p>0$ does the series $\sum}_{n=1}^{\infty}\frac{{n}^{p}}{{2}^{n}$ converge?
Converges for all $p>0.$
For which $r>0$ does the series $\sum}_{n=1}^{\infty}\frac{{r}^{{n}^{2}}}{{2}^{n}$ converge?
For which $r>0$ does the series $\sum}_{n=1}^{\infty}\frac{{2}^{n}}{{r}^{{n}^{2}}$ converge?
Converges for all $r>1.$ If $r>1$ then ${r}^{n}>4,$ say, once $n>\text{ln}\left(2\right)\text{/}\text{ln}\left(r\right)$ and then the series converges by limit comparison with a geometric series with ratio $1\text{/}2.$
Find all values of $p$ and $q$ such that $\sum}_{n=1}^{\infty}\frac{{n}^{p}}{{\left(n\text{!}\right)}^{q}$ converges.
Does $\sum}_{n=1}^{\infty}\frac{{\text{sin}}^{2}\left(nr\text{/}2\right)}{n$ converge or diverge? Explain.
The numerator is equal to $1$ when $n$ is odd and $0$ when $n$ is even, so the series can be rewritten $\sum}_{n=1}^{\infty}\frac{1}{2n+1},$ which diverges by limit comparison with the harmonic series.
Explain why, for each $n,$ at least one of $\{\left|\text{sin}\phantom{\rule{0.1em}{0ex}}n\right|,\left|\text{sin}\left(n+1\right)\right|\text{,...},\left|\text{sin}\phantom{\rule{0.1em}{0ex}}n+6\right|\}$ is larger than $1\text{/}2.$ Use this relation to test convergence of $\sum}_{n=1}^{\infty}\frac{\left|\text{sin}\phantom{\rule{0.1em}{0ex}}n\right|}{\sqrt{n}}.$
Suppose that ${a}_{n}\ge 0$ and ${b}_{n}\ge 0$ and that $\sum _{n=1}^{\infty}{a}^{2}{}_{n}$ and $\sum _{n=1}^{\infty}{b}^{2}{}_{n}$ converge. Prove that $\sum _{n=1}^{\infty}{a}_{n}{b}_{n}$ converges and $\sum _{n=1}^{\infty}{a}_{n}}\phantom{\rule{0.2em}{0ex}}{b}_{n}\le \frac{1}{2}\left({\displaystyle \sum _{n=1}^{\infty}{a}_{n}^{2}}+{\displaystyle \sum _{n=1}^{\infty}{b}_{n}^{2}}\right).$
${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ or ${a}^{2}+{b}^{2}\ge 2ab,$ so convergence follows from comparison of $2{a}_{n}{b}_{n}$ with ${a}^{2}{}_{n}+{b}^{2}{}_{n}.$ Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.
Does $\sum _{n=1}^{\infty}{2}^{\text{\u2212}\text{ln}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}n}$ converge? ( Hint: Write ${2}^{\text{ln}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}n}$ as a power of $\text{ln}\phantom{\rule{0.1em}{0ex}}n.)$
Does $\sum _{n=1}^{\infty}{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{\text{\u2212}\text{ln}\phantom{\rule{0.1em}{0ex}}n}$ converge? ( Hint: Use $t={e}^{\text{ln}\left(t\right)}$ to compare to a $p-\text{series}\text{.})$
${\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{\text{\u2212}\text{ln}\phantom{\rule{0.1em}{0ex}}n}={e}^{\text{\u2212}\text{ln}\left(n\right)\text{ln}\phantom{\rule{0.1em}{0ex}}\text{ln}\left(n\right)}.$ If $n$ is sufficiently large, then $\text{ln}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}n>2,$ so ${\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{\text{\u2212}\text{ln}\phantom{\rule{0.1em}{0ex}}n}<1\text{/}{n}^{2},$ and the series converges by comparison to a $p-\text{series}\text{.}$
Does $\sum _{n=2}^{\infty}{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}n\right)}^{\text{\u2212}\text{ln}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}n}$ converge? ( Hint: Compare ${a}_{n}$ to $1\text{/}n.)$
Show that if ${a}_{n}\ge 0$ and $\sum _{n=1}^{\infty}{a}_{n}$ converges, then $\sum _{n=1}^{\infty}{a}^{2}{}_{n}$ converges. If $\sum _{n=1}^{\infty}{a}^{2}{}_{n}$ converges, does $\sum _{n=1}^{\infty}{a}_{n}$ necessarily converge?
${a}_{n}\to 0,$ so ${a}^{2}{}_{n}\le \left|{a}_{n}\right|$ for large $n.$ Convergence follows from limit comparison. $\sum 1\text{/}{n}^{2}$ converges, but $\sum 1\text{/}n$ does not, so the fact that $\sum _{n=1}^{\infty}{a}^{2}{}_{n}$ converges does not imply that $\sum _{n=1}^{\infty}{a}_{n}$ converges.
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