Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms 
Sum this series: $$\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\ldots$$ upto $n$ terms.  

My approach:
$$(1-n^6)=(1-n^2)(1+n^2+n^4)\implies \dfrac{n}{1+n^2+n^4}=\dfrac{n(1-n^2)}{1-n^6}$$  
So, the above series can be written as $$\sum\limits_{i=1}^n \dfrac{i(1-i^2)}{1-i^6}$$  
I suppose that this can now be converted into integration which I cannot apparently. Please help. It would be better if the solution is not based upon integration but algebra.
 A: HINT:
As $\displaystyle1+n^2+n^4=(1+n^2)^2-n^2=(1+n+n^2)(1-n+n^2)$
$$\frac{2n}{1+n^2+n^4}=\frac{(1+n+n^2)-(1-n+n^2)}{(1+n+n^2)(1-n+n^2)}=\cdots$$
Again if $\displaystyle f(n)=\frac1{n^2-n+1}, f(n+1)=?$
So, we are dealing with a Telescoping series
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Lets $\ds{\mathcal{R} \equiv \braces{r}}$ the set of simple poles of $\ds{\pars{z^{4} + z^{2} + 1}^{-1}}$. Then,
\begin{align}
{k \over k^{4} + k^{2} + 1} & =
\sum_{r \in \mathcal{R}}{1 \over 4r^{3} + 2r}{k \over k - r}\ =\
\overbrace{\sum_{r \in \mathcal{R}}{1 \over 4r^{3} + 2r}}^{\ds{=\ 0}}\
+\
\sum_{r \in \mathcal{R}}{r \over 4r^{3} + 2r}{1 \over k - r}
\end{align}

\begin{align}
\color{#f00}{\sum_{k = 1}^{n}{k \over k^{4} + k^{2} + 1}} & =
\sum_{r \in \mathcal{R}}{r \over 4r^{3} + 2r}
\sum_{k = 0}^{n - 1}{1 \over k + 1 - r}\tag{1}
\end{align}

Also
\begin{align}
\sum_{k = 0}^{n - 1}{1 \over k + 1 - r} & =
\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 - r} - {1 \over k + n + 1 - r}}
\\[3mm] & =
n\sum_{k = 0}^{\infty}{1 \over \pars{k + n + 1 - r}\pars{k + 1 - r}} =
\Psi\pars{n + 1 - r} - \Psi\pars{1 - r}
\end{align}
where $\Psi$ is the Digamma function and we used a well known identity.

Then $\ds{\pars{~see\ expression\ \pars{1}~}}$
$$
\color{#f00}{\sum_{k = 1}^{n}{k \over k^{4} + k^{2} + 1}} =
\color{#f00}{\sum_{r \in \mathcal{R}}{r \over 4r^{3} + 2r}
\bracks{\Psi\pars{n + 1 - r} - \Psi\pars{1 - r}}}
$$


Note that the final result is expressed as a finite sum which only has four terms.

