According to Wolfram Alpha $$\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2} = \frac{\pi^2}{12}-\frac{\ln^22}{2}$$ I searched on Wikipedia and learnt that $$\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2} = \mathrm{Li}_2\left(\frac{1}{2}\right)$$ In general, the series is related to polylogarithm function $$\mathrm{Li}_k(z)= \sum_{n=1}^{\infty} \frac{z^n}{n^k}$$ However, I do not understand how exactly to use the polylogarithm function to evaluate $$\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2}$$ Could people provide me some assistance?


marked as duplicate by Jack D'Aurizio sequences-and-series Oct 11 '18 at 22:00

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    $\begingroup$ I am quite sure that your sum equals $\operatorname{Li}_2\left(\frac12\right)$ and not $\operatorname{Li}_2(2)$ as you claimed. $\endgroup$ – mrtaurho Oct 11 '18 at 21:47
  • $\begingroup$ Right, I will make the correction. $\endgroup$ – Larry Oct 11 '18 at 21:48

The third identity here https://en.wikipedia.org/wiki/Spence%27s_function#Identities is easily proved by integration \begin{eqnarray*} \mathrm{Li}_2(z)+\mathrm{Li}_2(1-z) = \frac{\pi^2}{6} - \ln(z) \ln(1-z). \end{eqnarray*} Now set $z=1/2$ and we have the result \begin{eqnarray*} \mathrm{Li}_2 \left( \frac{1}{2} \right) = \sum_{n=1}^{\infty} \frac{1}{n^2 2^n}= \frac{\pi^2}{12} - \frac{1}{2} (\ln(2))^2 . \end{eqnarray*}


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