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.
 A: 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.
A: Here are the definitions of the Riemann zeta function and Dirichlet eta function:
$$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}\cdots$$
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$
The following manipulation relates these two functions:
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\frac{1}{5^s}-\frac{1}{6^s}+\cdots$$
$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\cdots\right)$$
$$=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)-\frac{2}{2^s}\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right) $$
$$=\left(1-\frac{2}{2^s}\right)\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots\right)=(1-2^{1-s})\zeta(s) $$
In fact, since $\eta(s)$ has abscissa of convergence ${\rm Re}(s)>0$ whereas $\zeta(s)$ has abscissa of convergence ${\rm Re}(s)>1$, this can be used to provide an analytic continuation of $\zeta(s)$ to the so-called critical strip $0<{\rm Re}(s)<1$. You seek $\eta(2)$, and computing $\zeta(2)=\frac{\pi^2}{6}$ is known famously as the Basel problem. See the link for many solutions to the problem.
Thus $\eta(2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$.
A: One may derive using the residue theorem:
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+a^2} = \frac12 \left (\frac1{a^2}-\frac{\pi}{a} \operatorname{csch}{\pi a} \right ) $$
Take the limit as $a \to 0$.  Use the fact that
$$\operatorname{csch}{z} = \frac1{z} - \frac{z}{6} + O(z^3)$$
and the result that the desired sum is $\pi^2/12$ follows.
