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.

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, $$ \sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7} $$

Let us use the Abel-Plana formula $$ \sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{e^{2\pi t}-1} \, dt $$ for $f(n)=\dfrac{1}{1+n^2+n^4}$.

Since $ f(it)=f(-it) $ we obtain $$ \sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0). $$ By standard way we get $$ \int_0^\infty\frac{1}{1+x^2+x^4}=\frac{\pi \sqrt{3}}{6}. $$ Now $$ \sum_{i=1}^\infty f(n)=\sum_{i=0}^\infty f(n)-f(0)=\frac{\pi \sqrt{3}}{6}-\frac{1}{2}. $$