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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 $\sum \frac{1}{(n^2+1)^2}$?

flag	asked May 6 2012 at 21:48 David Feldman 11.4k●3●25●79

2

$$ \sum_{n = 0}^{\infty} (n^{2} + 1)^{-2} = \frac{-\pi^{2}}{4} + \frac{\pi^{2} coth (\pi)^{2}}{4} + \frac{1}{2} + \frac{\pi coth (\pi)}{4} $$ – Gerald Edgar May 6 2012 at 22:06

Interesting decimal... $\sum \frac{n^2-1}{(n^2+1)^2}=.453000$ – Gerald Edgar May 6 2012 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 2012 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 2012 at 22:33

Michael Renardy · Accepted Answer · 2012-05-06 23:34:11Z UTC

8

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

answered May 6 2012 at 23:34

Michael Renardy
4,880●1●10●17

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

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