Infinite sum $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ I am interested in the following result
\begin{equation*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}
\end{equation*}
I want to know the fundamentals. 
Thankful in advance.

Let $S:=\sum_{k=1}^\infty\,\frac{1}{(2k-1)^2}$.  Then, $\frac{\pi^2}{6}=\sum_{k=1}^\infty \,\frac{1}{k^2}=\sum_{r=0}^\infty \,\frac{S}{4^r}=\frac{4}{3}S$.  That is, $S=\frac{\pi^2}{8}$.  Ergo, $\sum_{n=1}^\infty\,\frac{(-1)^{n-1}}{n^2}=\sum_{k=1}^\infty\,\frac{1}{(2k-1)^2}-2\sum_{k=1}^\infty\,\frac{1}{k^2}=2\left(\frac{\pi^2}{8}\right)-\frac{\pi^2}{6}=\frac{\pi^2}{12}$.  Your answer is wrong.
