<s>[INST] 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.
 [/INST] As an alternative since
$$\frac{n}{n^4+n^2+1}\sim \frac1{n^3}$$
we can conclude that the series converges by limit comparison test with $\sum \frac1{n^3}$.
For the evaluation let consider
$$\frac{n}{n^{4}+n^{2}+1} = \frac{1}{2(n^{2}-n+1)}-\frac{1}{2(n^{2}+n+1)}$$
and use telescoping, indeed for $n=k+1$ we have $$\frac{1}{n^{2}-n+1}=\frac{1}{k^{2}+k+1}$$
and therefore
$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}=\frac12\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=$$
$$=\frac12\left(1+\sum_{n=2}^{\infty}\frac{1}{n^{2}-n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12\left(1+\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}-\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}\right)=\frac12$$
</s>