Sum of $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$ I came across these two series $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$$ and $$\sum_{n=1}^\infty \frac{(-1)^n(n^2-1)}{n^2+(n^2-1)^2}.$$ Does anyone recognize the sums of these series?
 A: They can be evaluated in terms of the digamma function, since:
$$ n^2 + (n^2-1)^2 = n^4 - n^2 + 1 = (n^2-\omega)(n^2-\omega^2), $$
with $\omega=\exp(2\pi i/3)$, and:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+\alpha}=\frac{-1+\pi\sqrt{\alpha}\coth(\pi\sqrt{\alpha})}{2\alpha},$$
$$\sum_{n=1}^{+\infty}\frac{(-1)^n}{n^2+\alpha}=\frac{-2+\pi\sqrt{\alpha}\coth(\pi\sqrt{\alpha}/2)-\pi\sqrt{\alpha}\tanh(\pi\sqrt{\alpha}/2)}{2\alpha}.$$
A: Using the Residue Theorem, we can show that, for sufficiently well-behaved functions $f(z)$:
$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \sum_{k=1}^m \mathrm{Res}_{z=z_k} \pi \csc{(\pi z)} f(z)$$
where the $z_k$ are the poles of $f$.  I will not show this for now, only that I note that $f(z) = (z^2 + (z^2-1)^2)^{-1}$ is sufficiently well-behaved so that we may use the above formula to evaluate the sum.
To find the $z_k$, we solve $ z^2 + (z^2-1)^2 = z^4-z^2+1 = 0$, or
$$z_{(\pm \pm )} = \pm \exp{\left (\pm i \frac{\pi}{6} \right )}$$
We now evaluate the residues associated with the poles and add them up.  The result (after some algebra) is
$$\sum_{k=1}^m \mathrm{Res}_{z=z_k} \pi \csc{(\pi z)} f(z) = \pi \csc{\left ( \frac{\pi}{3} \right )} \Im{\left [\exp{\left (-i \frac{\pi}{6} \right )} \csc{\left (\pi \exp{\left (i \frac{\pi}{6} \right )}\right )}\right ]} $$
The computation boils down to the evaluation of $\csc{\left (\pi \exp{\left (i \frac{\pi}{6} \right )}\right )}$.  Again, I will spare the reader for now.  The result is
$$S = \sum_{n=-\infty}^{\infty} (-1)^n f(n) = \frac{2 \pi}{\sqrt{3}} \frac{\sqrt{3} \cos{\left ( \sqrt{3} \frac{\pi}{2} \right )} \sinh{\left ( \frac{\pi}{2} \right )}+\sin{\left ( \sqrt{3} \frac{\pi}{2} \right )} \cosh{\left ( \frac{\pi}{2} \right )}}{\cosh{\pi} - \cos{(\sqrt{3} \pi)}}$$
However, this is not what we were asked for; rather, the result we seek is
$$\sum_{n=1}^{\infty} (-1)^n f(n) = \frac{1}{2} (S-1) $$
I verified this with a numerical and analytical result in Mathematica.
