The following series:
$$\sum_{k=0}^{\infty}\frac{k}{2^{k+1}\left ( k+1 \right )^2}$$
came up as an intermediate step of calculating an integral. The answer according to Wolfram is $\displaystyle \sum_{k=0}^{\infty}\frac{k}{2^{k+1}\left ( k+1 \right )^2}=\ln 2 +\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$.
Unfortunately , I have no idea how to attack this. Although I have some suspicions that it should be connected to some of the special functions and that we should differentiate. (i.e ${\rm Li}_n$)
Also, that zero index could easily be changed to $1$ since the fist summand is , obviously, zero. Can anyone help me attain the result. Both real analysis and complex analysis are welcomed...
$$f(x) = \sum_{k=0}^{\infty} \frac{k x^{k+1}}{(k+1)^2} $$
$$f'(x) = \sum_{k=0}^{\infty} \frac{k x^{k}}{k+1} = \sum_{k=0}^{\infty} x^k - \sum_{k=0}^{\infty} \frac{x^k}{k+1} = \frac1{1-x} + \frac{\log{(1-x)}}{x}$$
The sum is then
$$f \left ( \frac12 \right ) = \int_0^{1/2} dx \left [\frac1{1-x} + \frac{\log{(1-x)}}{x} \right ] = \log{2} - \operatorname{Li}_2\left ( \frac12 \right )$$
Use the fact that $\operatorname{Li}_2(1/2) = \pi^2/12 - \log^2{2}/2$ and the result follows.