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How do I calculate the sum of this series (studying for a test, not homework)?

$$\sum_{n=1}^\infty \frac{(-1)^n}{n2^{n+1}}$$

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If you know power series, you can look at $f(x):=\sum_{n=1}^{+\infty}\frac{(-1)^n x^n}n$. – Davide Giraudo Nov 2 '11 at 15:52
@Davide Can you expand on that? – Caleb Jares Nov 2 '11 at 15:57
You perhaps know that $\frac{1}{1+x} = \sum_{n \geq 0} (-1)^n x^n$. What will you get if you integrate both sides w.r.t. $x$ (and set $x = 1/2$)? – Srivatsan Nov 2 '11 at 15:58
To see how anything is formatted, right click and choose Show Source. – Ross Millikan Nov 2 '11 at 16:01
@Ross Thanks for the tip. I always wondered how to hitch a ride. – TheBirdistheWord Nov 3 '11 at 16:01
up vote 13 down vote accepted

A useful heuristic is to combine as much as possible into $n$th powers: $$\sum_{n=1}^{\infty} \frac1{2n}\left(\frac{-1}{2}\right)^n$$ which is $$\frac12 \sum_{n=1}^\infty \frac{x^n}{n}\quad \text{with }x=-1/2$$ If we don't immediately recognize $\sum \frac{x^n}{n}$, differentiate it symbolically to get $\sum_{n=0}^\infty x^n$ which is a geometric series with sum $\frac1{1-x}$ and then integrate that to get $-\log(1-x)$ (with constant of integration selected to make the 0th order terms match).

So $\frac 12 \sum_{n=1}^\infty \frac{x^n}{n} = -\frac 12\log(1-x)$, and thus the sought answer is $-\frac12\log(1+\frac 12) = -\frac 12\log \frac{3}{2}$.

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The Taylor series for $\log(1+x)$ is $$ \log(1+x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}{k} $$ Pluging in $x=\frac{1}{2}$, we get $$ \log\left(\frac{3}{2}\right)=\sum_{k=1}^\infty(-1)^{k-1}\frac{1}{k2^k} $$ Multiplying by $-\frac{1}{2}$ yields $$ -\frac{1}{2}\log\left(\frac{3}{2}\right)=\sum_{k=1}^\infty(-1)^{k}\frac{1}{k2^{k+1}} $$

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Congrats on the 10k :-) – Asaf Karagila Nov 13 '11 at 22:16
@Asaf: Thanks! :-) – robjohn Nov 13 '11 at 22:24

$$ \begin{eqnarray} \sum_{n=1}^\infty \frac{(-1)^n}{n2^{n+1}} &=& \left.\frac12\sum_{n=1}^\infty \frac{q^n}n\right|_{q=-1/2}\\ &=& \left.\frac12\sum_{n=1}^\infty \int_0^qt^{n-1}\mathrm dt\right|_{q=-1/2} \\ &=& \left.\frac12\int_0^q\sum_{n=1}^\infty t^{n-1}\mathrm dt\right|_{q=-1/2} \\ &=& \left.\frac12\int_0^q\frac1{1-t}\mathrm dt\right|_{q=-1/2} \\ &=& \left.\frac12\big[-\log(1-t)\big]_0^q\right|_{q=-1/2} \\ &=& \left.\frac12\left(-\log(1-q)\right)\right|_{q=-1/2} \\ &=& -\frac12\log\frac32\;. \end{eqnarray} $$

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