Sum of $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$

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: $$\sum_{n=1}^{+\infty}\frac{1}{n^2+\alpha}=\frac{-1+\pi\sqrt{\alpha}\coth(\pi\sqrt{\alpha})}{2\alpha},$$ $$\sum_{n=1}^{+\infty}\frac{(-1)^n}{n^2+\alpha}=\frac{-2+\pi\sqrt{\alpha}\coth(\pi\sqrt{\alpha}/2)-\pi\sqrt{\alpha}\tanh(\pi\sqrt{\alpha}/2)}{2\alpha}.$$

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.