Check the convergence of $\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$ and if it converges find the sum Check whether the series converges or not. If yes, find the sum of the series $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
My Efforts
Observe $n^3<n^4+n^2+1$
$\therefore$
$\frac{1}{n^4+n^2+n+1}<\frac{1}{n^3}$ which further implies that $\frac{n}{n^4+n^2+n+1}<\frac{n}{n^3}=\frac{1}{n^2}$
Since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, we conclude by comparison test that the series converges. 
I am not able to find the sum of this series? Any hints or directions will be appreciated.

We have
$$n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$
$$((n+1)^2-(n+1)+1)=(n^2+2n+1-n-1+1)=(n^2+n+1)$$
