Help in finding the sum of the series $$\sum_{n=1}^\infty \frac{1}{n^4+n^2+1}$$
I tried breaking into factors but it is not telescoping.
$$\frac {1}{(n^2+n+1)(n^2-n+1)} = \frac {1}{2n} \left(\frac {1}{n^2-n+1} - \frac {1}{n^2+n+1}\right)$$

Let $\omega=\exp\frac{2\pi i}{3}$. Then:
$$\sum_{n\geq 1}\frac{1}{n^4+n^2+1}=\frac{1}{i\sqrt{3}}\left(\sum_{n\geq 1}\frac{1}{n^2-\omega}-\sum_{n\geq 1}\frac{1}{n^2-\omega^2}\right)=\color{red}{-\frac{1}{2}+\frac{\pi\sqrt{3}}{6}\tanh\frac{\pi\sqrt{3}}{2}}$$
since we can deal with series like the ones appearing in the middle term through the logarithmic derivative of the Weierstrass product for the hyperbolic sine function. In particular, the identity:
$$ \sum_{n\geq 0}\frac{1}{n^2+z^2}=\frac{1+\pi z\coth(\pi z)}{2z^2}$$
is well-known.
