<< Chapter < Page | Chapter >> Page > |
Use the fact that
to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is $3\phantom{\rule{0.1em}{0ex}}\text{ln}\left(2\right)\text{/}2.$
Let
Since $\sum _{n=1}^{\infty}{a}_{n}}=\text{ln}(2),$ by the algebraic properties of convergent series,
Now introduce the series $\sum _{n=1}^{\infty}{b}_{n}$ such that for all $n\ge 1,$ ${b}_{2n-1}=0$ and ${b}_{2n}={a}_{n}\text{/}2.$ Then
Then using the algebraic limit properties of convergent series, since $\sum _{n=1}^{\infty}{a}_{n}$ and $\sum _{n=1}^{\infty}{b}_{n}$ converge, the series $\sum _{n=1}^{\infty}({a}_{n}+{b}_{n})$ converges and
Now adding the corresponding terms, ${a}_{n}$ and ${b}_{n},$ we see that
We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since $\sum _{n=1}^{\infty}({a}_{n}+{b}_{n})}=3\phantom{\rule{0.1em}{0ex}}\text{ln}(2)\text{/}2,$ we conclude that
Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.
State whether each of the following series converges absolutely, conditionally, or not at all.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{n}{n+3$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{\sqrt{n}+1}{\sqrt{n}+3$
Does not converge by divergence test. Terms do not tend to zero.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{1}{\sqrt{n+3}$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{\sqrt{n+3}}{n$
Converges conditionally by alternating series test, since $\sqrt{n+3}\text{/}n$ is decreasing. Does not converge absolutely by comparison with p -series, $p=1\text{/}2.$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{1}{n\text{!}$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{{3}^{n}}{n\text{!}$
Converges absolutely by limit comparison to ${3}^{n}\text{/}{4}^{n},$ for example.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\left(\frac{n-1}{n}\right)}^{n$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\left(\frac{n+1}{n}\right)}^{n$
Diverges by divergence test since $\underset{n\to \infty}{\text{lim}}|{a}_{n}|=e.$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\text{sin}}^{2}n$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\text{cos}}^{2}n$
Does not converge. Terms do not tend to zero.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\text{sin}}^{2}\left(1\text{/}n\right)$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{\text{cos}}^{2}\left(1\text{/}n\right)$
$\underset{n\to \infty}{\text{lim}}{\text{cos}}^{2}(1\text{/}n)=1.$ Diverges by divergence test.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\text{ln}\left(1\text{/}n\right)$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\text{ln}\left(1+\frac{1}{n}\right)$
Converges by alternating series test.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{{n}^{2}}{1+{n}^{4}$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{{n}^{e}}{1+{n}^{\pi}$
Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p -series, $p=\pi -e$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{2}^{1\text{/}n$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}{n}^{1\text{/}n$
Diverges; terms do not tend to zero.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n}\left(1-{n}^{1\text{/}n}\right)$ ( Hint: ${n}^{1\text{/}n}\approx 1+\text{ln}(n)\text{/}n$ for large $n.)$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}n\left(1-\text{cos}\left(\frac{1}{n}\right)\right)$ ( Hint: $\text{cos}(1\text{/}n)\approx 1-1\text{/}{n}^{2}$ for large $n.)$
Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)$ ( Hint: Rationalize the numerator.)
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$ ( Hint: Cross-multiply then rationalize numerator.)
Converges absolutely by limit comparison with p -series, $p=3\text{/}2,$ after applying the hint.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\left(\text{ln}\left(n+1\right)-\text{ln}\phantom{\rule{0.2em}{0ex}}n\right)$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}n\left({\text{tan}}^{\mathrm{-1}}\left(n+1\right)-{\text{tan}}^{\mathrm{-1}}n\right)$ ( Hint: Use Mean Value Theorem.)
Converges by alternating series test since $n({\text{tan}}^{\mathrm{-1}}(n+1)\text{\u2212}{\text{tan}}^{\mathrm{-1}}n)$ is decreasing to zero for large $n.$ Does not converge absolutely by limit comparison with harmonic series after applying hint.
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\left({\left(n+1\right)}^{2}-{n}^{2}\right)$
$\sum}_{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)$
Converges absolutely, since ${a}_{n}=\frac{1}{n}-\frac{1}{n+1}$ are terms of a telescoping series.
Notification Switch
Would you like to follow the 'Calculus volume 2' conversation and receive update notifications?