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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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    @Ross Thanks for the tip. I always wondered how to hitch a ride.2011-11-03

3 Answers 3

13

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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    @Asaf: Thanks! :-)2011-11-13
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$ \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} $