\(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}\)
In the event you're interested, the series consists of the negative of the sum of the squares of the reciprocals of the ODDS, and the sum of the squares of the reciprocals of the EVENS.
\(\displaystyle -1+\frac{1}{4}-\frac{1}{9}+\frac{1}{16}-\frac{1}{25}+...................\)
\(\displaystyle \left(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+......\right)-\left(1+\frac{1}{9}+\frac{1}{25}+............\right)\)
\(\displaystyle \sum_{n=1}^{\infty}\frac{-1}{(2n-1)^{2}}=\frac{-{\pi}^{2}}{8}\)
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^{2}}=\frac{{\pi}^{2}}{24}\)
\(\displaystyle \frac{{\pi}^{2}}{24}-\frac{{\pi}^{2}}{8}=\frac{-{\pi}^{2}}{12}\)