I know that the infinite sum of the reciprocals of squares converges to $\pi^2/6$. Interested by this, I looked at a different sum. It is similar to the previously mentioned series, but it alternates signs: $\sum_{i=1}^n \frac{(-1)^{i+1}}{i^2}$. I tried adding up the first several terms but I could not identify any interesting convergence (up to $\ n=14$ the sum is 0.82009731292). Is it possible that the series does not converge?
Thanks.
Mathematica says the sum is $\pi^2/12,$ and this is not not hard to prove - break the sum into the even and the odd parts, and transform into the sum of reciprocal squares.
Thanks, Igor. I think I figured it out. You take the original sum which converges to $\pi^2/6$. The even part is a half squared plus a fourth squared plus a sixth squared, etc. Factoring out a fourth shows that this even part sum is one fourth of the total sum. So, the odd part is 3/4 of the sum. 3/4 - 1/4 is a half, so the series converges to $\pi^2/12$.
Again, thanks.
A nice way to calculate this sum uses residue theory. Drawing a contour C around the origin on the complex plane, we obtain $$\int_C \frac{\pi z \csc \pi z}{z^2}dz=2\pi i\ \text{Res}_{z=0}\frac{\pi z \csc \pi z}{z^2}-2\pi i \sum_{n=-m\\n\neq0}^m\frac{(-1)^{n+1}}{z^2}$$ As we extend C out to infinity, the integral on the left goes to zero, while the summation limits on the right go to infinity. This sum is then twice the alternating series $\eta(2)$ mentioned in the question, and is equal the residue at zero. $$\text{Res}_{z=0}\frac{\pi z \csc \pi z}{z^2}=\frac{\pi^2}{6}=2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{z^2}=2\eta(2)$$ Therefore, $\eta(2)=\pi^2/12$.
