<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?
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\begin{align}
&\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1}
=
\sum_{n = 1}^{\infty}{n \over \pars{n^{2} + 1/2}^{2} + 3/4}
\\[3mm]&=
\sum_{n = 1}^{\infty}{n \over \bracks{n^{2} -\pars{-1/2 - \sqrt{3}\,\ic/2}}
                         \bracks{n^{2} -\pars{-1/2 + \sqrt{3}\,\ic/2}}}
=
\sum_{n = 1}^{\infty}{n \over \pars{n^{2} - \xi^{2}}\pars{n^{2} - {\xi^{*}}^{2}}}
\end{align}
where $\xi^{2} \equiv \pars{-1 - \root{3}\,\ic}/2 = \expo{4\pi\ic/3}$

\begin{align}
&\sum_{n = 1}^{N}{n \over n^{4} + n^{2} + 1}
=
{1 \over \xi^{2} - {\xi^{*}}^{2}}\sum_{n = 1}^{N}\pars{%
{n \over n^{2} - \xi^{2}} - {n \over n^{2} - {\xi^{*}}^{2}}}
=
{1 \over 2\ic\Im\pars{\xi^{2}}}\,2\ic\Im\sum_{n = 1}^{N}
{n \over n^{2} - \xi^{2}}
\\[3mm]&=
-\,{2\root{3} \over 3}\Im\bracks{{1 \over 2}\sum_{n = 1}^{N}
\pars{{1 \over n - \xi} + {1 \over n + \xi}}}
=
-\,{\root{3} \over 3}\Im\sum_{n = 1}^{N}
\pars{{1 \over n + \xi} + {1 \over n - \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 1}^{N}{1 \over n - \xi^{*}}}
\end{align}

\begin{align}
&\xi = \expo{2\pi\ic/3} = \cos\pars{2\pi \over 3} + \sin\pars{2\pi \over 3}\ic
=
-\,{1 \over 2} + {\root{3} \over 2}\,\ic
\\[3mm]&
\mbox{Notice that}\ \xi^{*} = -\,{1 \over 2} - {\root{3} \over 2}\,\ic 
= -1 - \pars{-\,{1 \over 2} + {\root{3} \over 2}\,\ic} = - 1 - \xi
\end{align}

\begin{align}
&\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1}
=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 1}^{N}{1 \over n + 1 + \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 2}^{N + 1}{1 \over n + \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
{1 \over 1 + \xi } + \sum_{n = 2}^{N}{1 \over n + \xi}
-
\sum_{n = 2}^{N}{1 \over n + \xi} - {1 \over N + 1 + \xi}}
=
-\,{\root{3} \over 3}\Im\pars{1 \over 1 + \xi }
\\[3mm]&=
-\,{\root{3} \over 3}\Im\bracks{1 \over \pars{1 + \root{3}\ic}/2}
=
-\,{\root{3} \over 6}\Im\pars{1 - \root{3}\ic} = {1 \over 2}
\end{align}

$$\color{#0000ff}{\large%
\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1} = {1 \over 2}}
$$
</s>