<s>[INST] 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?
 [/INST] HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)$
and $(n^2+1+n)-(n^2+1-n)=2n$
$$\frac n{n^4+n^2+1}=\frac12\left(\frac{2n}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac{(n^2+1+n)-(n^2+1-n)}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac1{n^2-n+1}-\frac1{n^2+n+1}\right)$$
Also observe that $: (n+1)^2-(n+1)+1=n^2+n+1$ inviting cancellations 
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