Evaluating $ \sum\frac{1}{1+n^2+n^4} $ How to evaluate following expression?
$$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$
I doubt it is a telescopic Sum.

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then:
$$ n^4+n^2+1 = (n^2-\omega)(n^2-\omega^2), $$
so:
$$\frac{1}{1+n^2+n^4}=\frac{1}{i\sqrt{3}}\left(\frac{1}{n^2-\omega}-\frac{1}{n^2-\omega^2}\right)$$
and:
$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{\sqrt{3}}\Im\sum_{n=1}^{+\infty}\frac{1}{n^2-\omega}$$
can be computed through the identity:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2+a}=\frac{-1+\pi\sqrt{a}\coth(\pi\sqrt{a})}{2a}$$
that follows from considering the logarithmic derivative of the Weierstrass product for the $\sinh$ function. By putting all together, we have:

$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{6}\left(-3+\pi\sqrt{3}\tanh\frac{\pi\sqrt{3}}{2}\right)$$

as stated by WA.
