Sum of the series $\sum_{n=1}^\infty \frac{(-1)^n}{n2^{n+1}}$ How do I calculate the sum of this series (studying for a test, not homework)?
$$\sum_{n=1}^\infty \frac{(-1)^n}{n2^{n+1}}$$
 A: $$
\begin{eqnarray}
\sum_{n=1}^\infty \frac{(-1)^n}{n2^{n+1}}
&=&
\left.\frac12\sum_{n=1}^\infty \frac{q^n}n\right|_{q=-1/2}\\
&=&
\left.\frac12\sum_{n=1}^\infty \int_0^qt^{n-1}\mathrm dt\right|_{q=-1/2}
\\
&=&
\left.\frac12\int_0^q\sum_{n=1}^\infty t^{n-1}\mathrm dt\right|_{q=-1/2}
\\
&=&
\left.\frac12\int_0^q\frac1{1-t}\mathrm dt\right|_{q=-1/2}
\\
&=&
\left.\frac12\big[-\log(1-t)\big]_0^q\right|_{q=-1/2}
\\
&=&
\left.\frac12\left(-\log(1-q)\right)\right|_{q=-1/2}
\\
&=&
-\frac12\log\frac32\;.
\end{eqnarray}
$$
A: 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 $-\frac12\log(1+\frac 12) = -\frac 12\log \frac{3}{2}$.
A: 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
$$
-\frac{1}{2}\log\left(\frac{3}{2}\right)=\sum_{k=1}^\infty(-1)^{k}\frac{1}{k2^{k+1}}
$$
