\(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{2}}=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+.................\)
Now, how in depth must you go?. Can we play off of other already known summations?. If not, we can try a more in-depth route.
We can use the well known \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}\)
Break them up in to odds and evens:
\(\displaystyle \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+....=\sum_{k=1}^{\infty}\frac{1}{(2k)^{2}}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{1}{4}\cdot\frac{{\pi}^{2}}{6}=\frac{{\pi}^{2}}{24}\)
Since these were all negatives, we can write \(\displaystyle \frac{{-\pi}^{2}}{24}\)
Now, the odds:
\(\displaystyle 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+........ = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}=\frac{3}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{8}\)
\(\displaystyle \frac{{\pi}^{2}}{8}-\frac{{\pi}^{2}}{24}=\frac{{\pi}^{2}}{12}\)
I played off of already well-known identities. The actual derivations of these can be involved.
For instance, to find the latter summation, one can note that \(\displaystyle \int_{0}^{1}\frac{sin^{-1}(t)}{\sqrt{1-t^{2}}}dt=\frac{{\pi}^{2}}{8}\)
This, combined with various expansions and recursions will give us the result.
