<s>[INST] 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.
 [/INST] 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 $\eta(2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$.
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