Evaluate $\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$ I am trying to re-learn some basic math and I realize I have forgotten most of it.

Evaluate $$\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$$

Call the terms $S_n$ and the total sum $S$.
$$S_n < \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1} = S <  \infty$$
$$S_n = \frac{n}{n^4+n^2+1} = \frac{n}{(n^2+1)^2-1}$$
It has been more than a few years since I did these things. 
I would like a hint about what method I should try to look for?
Thanks.

HINT:
$$n^4+n^2+1=(n^2-n+1)(n^2+n+1)$$
Write $2n$ as $$n^2+n+1-(n^2-n+1)$$
Observe that if $f(m)=m^2-m+1, f(m+1)=?$ 
which immediately reminds me of Telescoping Series.
