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Evaluate the series:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} $$

We use the formula \[\eta(s)=\bigl({1-2^{1-s}}\bigr)\,\zeta(s)\,,\quad \Re({s})>0\,,\] where \[\eta(s)=\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}\,,\quad \Re({s})>0\,,\] is the Dirichlet eta function and \(\zeta\) is Riemann's \(\zeta\) function. So
\begin{align*}
\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2}&=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}\\
&=\eta(2)\\
&=\bigl({1-2^{1-2}}\bigr)\,\zeta(2)\\
&=\frac{1}{2}\,\frac{\pi^2}{6}\\
&=\frac{\pi^2}{12}\,.\;{(\dagger)}
\end{align*}



\((\dagger)\) It is well known that \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}\,.\)