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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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marked as duplicate by Did sequences-and-series Sep 28 '18 at 7:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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

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  • $\begingroup$ @Hawk, First impression : for finite $n,$ definite integral won't work $\endgroup$ – lab bhattacharjee Feb 26 '14 at 11:50
  • $\begingroup$ 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? $\endgroup$ – Hawk Feb 26 '14 at 11:52
  • $\begingroup$ This is beautiful and elegant. (+1) $\endgroup$ – Ron Gordon Feb 26 '14 at 20:23
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

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