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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.

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  • $\begingroup$ The sum expression was wrong, so I corrected it. The alternating series converges, since the series converges absolutely. Not sure if this question will be well received. $\endgroup$
    – theHigherGeometer
    Commented May 21, 2015 at 3:22

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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.

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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.

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  • $\begingroup$ You should mark this answer as the solution. Thanks for explaining :) $\endgroup$
    – Rufflewind
    Commented Jan 7, 2016 at 19:37

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