# Evaluating $\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2}$ [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 11 '18 at 22:00

• I am quite sure that your sum equals $\operatorname{Li}_2\left(\frac12\right)$ and not $\operatorname{Li}_2(2)$ as you claimed. – mrtaurho Oct 11 '18 at 21:47
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 $$\begin{eqnarray*} \mathrm{Li}_2 \left( \frac{1}{2} \right) = \sum_{n=1}^{\infty} \frac{1}{n^2 2^n}= \frac{\pi^2}{12} - \frac{1}{2} (\ln(2))^2 . \end{eqnarray*}$$