Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}$ is convergent as well.
But I find it hard to calculate the sum. can you give me some hints?

Easiest and fastest solution:
Basically, try small values of $n$ and you'll see a pattern:
$\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+\frac{4}{4^4+4^2+1}\dots$
$\implies \frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\frac{4}{273}$
$\implies \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)+\frac{1}{4}\left(\frac{2}{3}-\frac{2}{7}\right)+\frac{1}{6}\left(\frac{3}{7}-\frac{3}{13}\right)+\frac{1}{8}\left(\frac{4}{13}-\frac{4}{21}\right)+\dots$
$\implies \frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{14}+\frac{1}{14}-\frac{1}{26}+\frac{1}{26}-\frac{1}{42}+\dots$
Notice that we can just cancel out every single term except for the original one, that is $\boxed{\frac{1}{2}}$ and we're done!
