sum of an alternating series How to evaluate the series below ?
$$
\sum_{n=0}^{\infty}\left(-1\right)^{n}\,{2n+1 \over \left(2n+1\right)^{2} + x^{2}}
$$
Can we reexpress it in term of an elementary function ?.$\,$
By the way, someone told me that it is equal to
$$
{\pi \over 4}\,{\rm sech}\left(\pi x \over 2\right)
$$
 A: You can use residue theory.  For example, one may show that
$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = -\sum_k \operatorname*{Res}_{z=z_k} \left [\pi \, \csc{(\pi z)} \, f(z)\right]$$
where the $z_k$ are the non-integral poles of $f$.
In this case, first note that the sum in question is
$$\frac12 \sum_{n=-\infty}^{\infty} (-1)^n \frac{2 n+1}{(2 n+1)^2+x^2}$$
$$f(z) = \frac{2 z+1}{(2 z+1)^2+x^2} $$
The poles of $f$ are at
$$z_{\pm} = -\frac12 \pm i \frac{x}{2}$$
Note that
$$\csc{\left (\frac{\pi}{2} \pm i \frac{\pi}{2} x \right )} = \operatorname*{sech}{\left ( \frac{\pi}{2} x\right )}$$
so that
$$-\operatorname*{Res}_{z=z_{\pm}} \left [\pi \, \csc{(\pi z)} \frac{2 z+1}{(2 z+1)^2+x^2} \right ] = \frac{\pi}{4} \operatorname*{sech}{\left ( \frac{\pi}{2} x\right )}$$
Therefore the sum over all integers is twice this (the residues of each of the poles is equal to the above), but since the desired sum is half of the sum over all integers, we have
$$\sum_{n=0}^{\infty} (-1)^n \frac{2 n+1}{(2 n+1)^2+x^2} = \frac{\pi}{4} \operatorname*{sech}{\left ( \frac{\pi}{2} x\right )}$$
