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.

First, we have
$$
\begin{align}
\frac1{z^4+z^2+1}
&=\frac1{12}\left(
\frac{-3-i\sqrt3}{z-e^{\pi i/3}}
+\frac{3+i\sqrt3}{z-e^{4\pi i/3}}
+\frac{3-i\sqrt3}{z-e^{2\pi i/3}}
+\frac{-3+i\sqrt3}{z-e^{5\pi i/3}}
\right)\tag{1}
\end{align}
$$
Let $\gamma$ be the rectangle
$$
[-1-i,1-i]\cup[1-i,1+i]\cup[1+i,-1+i]\cup[-1+i,-1-i]\tag{2}
$$
then the integral
$$
\frac1{2\pi i}\int_{(n+\frac12)\gamma}\frac{\pi\cot(\pi z)}{z^4+z^2+1}\,\mathrm{d}z\tag{3}
$$
tends to $0$ since along the horizontal paths, $|\pi\cot(\pi z)|\to\pi$ and along the vertical paths, $|\pi\cot(\pi z)|\lt\pi$.
Since $\pi\cot(\pi z)$ has residue $1$ at each integer, we get that $(3)$ is the sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ plus
$$
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}\tag{4}
$$
The sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ is
$$
-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\tag{5}
$$
Since the sum of $(4)$ and $(5)$ is $0$, we have
$$
\begin{align}
1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}
&=4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\\
&=-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\tan\left(\pi\tfrac{i\sqrt3}2\right)\right)\\
&=\tfrac{\pi\sqrt3}3\tanh\left(\pi\tfrac{\sqrt3}2\right)\tag{6}
\end{align}
$$
Therefore,
$$
\sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7}
$$
