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.

Notice, use partial fractions as follows  $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$$$=\sum_{n=1}^{\infty}\frac{n}{(n^2-n+1)(n^2+n+1)}$$
$$=\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$
$$=\frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\ldots +\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\right)$$
$$=\frac 12\lim_{n\to \infty}\left(1-\frac{1}{n^2+n+1}\right)$$
$$=\frac 12\left(1-0\right)=\color{red}{\frac 12}$$
