Please help me find in closed form a value for

$$ \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n^2+a^2} $$

Bearing in mind there is no "$x$" term, I would assume the solution involves instituting some form of $x^n$ and letting $S=f(1)$. I've tried doing so, along with taking derivatives of the function, hoping to somehow turn the series into a DE, so that the original function can be obtained. I've worked on it for hours, I'm baffled. Any help is appreciated.

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    $\begingroup$ Do you have a reason to think that this series has a closed form expression? $\endgroup$ – davidlowryduda Mar 19 '13 at 8:13
  • $\begingroup$ This may be done with Fourier series as proposed by muzzlator as shown in this answer (or with residus as shown by Argon). See $\frac 1{\sin(\pi z)}$ where you'll have to replace $z$ by $ia$. $\endgroup$ – Raymond Manzoni Mar 19 '13 at 8:36

Series like these may be evaluated by the Residue Theorem:

$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = -\sum_k \text{Res}_{z=z_k}[ \pi \csc{(\pi z)} f(z)]$$

where the $z_k$ are poles of $f$ away from the real line.

In your case

$$f(z) = \frac{1}{z^2+a^2}$$

The poles of $f$ are at $z_{\pm} = \pm i a$. Then the sum on the right is

$$-\pi \left( \frac{-i\, \text{csch}{(\pi a)}}{i 2 a} + \frac{i\, \text{csch}{(\pi a)}}{-i 2 a}\right ) $$


$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2} = \frac{\pi}{a} \text{csch}{(\pi a)}$$


$$\sum_{n=0}^{\infty} \frac{(-1)^n}{n^2+a^2} = \frac{1}{2} \left ( \frac{\pi}{a} \text{csch}{(\pi a)} + \frac{1}{a^2}\right)$$

  • $\begingroup$ That's really cool. I had heard of residue theory being used to calculate sums before but I have never seen it actually done $\endgroup$ – muzzlator Mar 19 '13 at 8:24
  • $\begingroup$ @muzzlator: in cases like this, it is very simple to apply. $\endgroup$ – Ron Gordon Mar 19 '13 at 8:26
  • $\begingroup$ Thank you much. You can't imagine the frustration I've had this past 12 hours. I would have never guessed... Thanks again! $\endgroup$ – Gerg Mar 19 '13 at 8:47
  • $\begingroup$ @Gerg: I am glad this helped you, and I hope you got the point for future, similar problems. Please remember to accept the solution you like best. $\endgroup$ – Ron Gordon Mar 19 '13 at 8:49
  • $\begingroup$ @RonGordon I don't know the residue theorem in this form, do you have a reference to this result please? $\endgroup$ – user63181 Mar 19 '13 at 9:22

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