# $\sum \frac{n^2-1}{(n^2+1)^2}=\frac{1}{2}(1-\frac{\pi^2}{\sinh(\pi)^2})$

The following series evaluation

$\sum \frac{n^2-1}{(n^2+1)^2}=\frac{1}{2}(1-\frac{\pi^2}{\sinh(\pi)^2})$

seems attractive to me, and has a proof related to the evaluation of $\zeta(2)$.

Does this identity (and/or it's many variants) occur in the literature? If so, in what context?

Also, does there exist a closed form for $\sum \frac{1}{(n^2+1)^2}$?

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$$\sum_{n = 0}^{\infty} (n^{2} + 1)^{-2} = \frac{-\pi^{2}}{4} + \frac{\pi^{2} coth (\pi)^{2}}{4} + \frac{1}{2} + \frac{\pi coth (\pi)}{4}$$ –  Gerald Edgar May 6 '12 at 22:06
Interesting decimal... $\sum \frac{n^2-1}{(n^2+1)^2}=.453000$ –  Gerald Edgar May 6 '12 at 22:25
Re: .453000: Yes! But I don't see the deeper meaning, even after looking at the continued fraction. –  David Feldman May 6 '12 at 22:32
Thanks to you and Maple, one also gets an evaluation of $\sum\frac{1}{n^2+1}$. I'm surprised it has one. –  David Feldman May 6 '12 at 22:33

This is a special case of $$\sum_{n=-\infty}^\infty f(n)=-\sum_{poles\ of\ f} Res(\pi\cot(\pi z)f(z)).$$ The formula is true for rational functions for which the series converges. Proofs can be found in complex analysis texts