# 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?

• This is a hit. A general method to calculate such sums of series is the calculation of certain contour integral. Also the Maple command $$sum(n/(n^4+n^2+1), n = 1 .. infinity)$$ outputs $1/2.$ Nov 18, 2013 at 15:58

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

• @Mohamez, The general technique for separating polynomial quotients like this is en.wikipedia.org/wiki/Partial_fraction_decomposition , which is a relatively simple and useful technique that only uses algebra so I strongly recommend learning it. Also note that the "cancellation" is only valid because the limit of the sequence is zero, because arbitrarily grouping the terms in an infinite sequence can give contradictory results. Logically you cancel the terms in the finite sequence, and then take the limit to $\infty$. Nov 18, 2013 at 16:24
• Jul 6, 2019 at 4:53

Use this equation$$\frac{2n}{n^4+n^2+1}=\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}.$$


\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}

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

Similar to Felix Marin's solution but using generalized harmonic numbers and partial sums $$\frac{n}{n^{4}+n^{2}+1}=\frac{1}{(a-b)}\Big(\frac{n}{n^2+b}-\frac{n}{n^2+a}\Big)$$ $$\sum_{n=1}^p \frac{n}{n^2+c}=\frac{1}{2} \left(H_{p-i \sqrt{c}}+H_{p+i \sqrt{c}}-H_{-i \sqrt{c}}-H_{i\sqrt{c}}\right)$$ Using the asymptotics of harmonic numbers $$\sum_{n=1}^p \frac{n}{n^2+c}=\left(-\frac{1}{2} H_{-i \sqrt{c}}-\frac{1}{2} H_{i \sqrt{c}}+\log \left({p}\right)+\gamma \right)+\frac{1}{2 p}+\frac{\frac{c}{2}-\frac{1}{12}}{p^2}+O\left(\frac{1}{p^3}\right)$$ Using it twice and continuing with Taylor expansions, wend with $$S_p=\sum_{n=1}^p \frac{n}{n^{4}+n^{2}+1}=\frac{1}{2}-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.

• I get $S_p=\frac12-\frac1{2p^2+2p+2}$, so I concur.
– robjohn
Jul 6, 2019 at 13:13
• @robjohn. This is the exact result Jul 6, 2019 at 14:56

This is just a telescoping series: \begin{align} \sum_{n=1}^\infty\frac{n}{n^4+n^2+1} &=\frac12\sum_{n=1}^\infty\left(\frac1{(n-1)n+1}-\frac1{n(n+1)+1}\right)\\ &=\frac12 \end{align} which is essentially what lab bhattacharjee says, but writing $$n^2-n+1=(n-1)n+1$$ and $$n^2+n+1=n(n+1)+1$$ makes this a bit simpler to see.

Easiest and fastest solution:

Basically, try small values of $$n$$ and you'll see a pattern:

$$\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+\frac{4}{4^4+4^2+1}\dots$$

$$\implies \frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\frac{4}{273}$$

$$\implies \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)+\frac{1}{4}\left(\frac{2}{3}-\frac{2}{7}\right)+\frac{1}{6}\left(\frac{3}{7}-\frac{3}{13}\right)+\frac{1}{8}\left(\frac{4}{13}-\frac{4}{21}\right)+\dots$$

$$\implies \frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{14}+\frac{1}{14}-\frac{1}{26}+\frac{1}{26}-\frac{1}{42}+\dots$$

Notice that we can just cancel out every single term except for the original one, that is $$\boxed{\frac{1}{2}}$$ and we're done!