# Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms [duplicate]

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.

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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

• @Hawk, First impression : for finite $n,$ definite integral won't work – lab bhattacharjee Feb 26 '14 at 11:50
• Yes, I missed that, if it were for infinite limit, how would that work? And, are there some other way in which my approach would have worked out? – Hawk Feb 26 '14 at 11:52
• This is beautiful and elegant. (+1) – Ron Gordon Feb 26 '14 at 20:23


\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.