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?
$\endgroup$ 16 Answers
$\begingroup$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
$\endgroup$ 2 $\begingroup$Use this equation$$\frac{2n}{n^4+n^2+1}=\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}.$$
$\endgroup$ $\begingroup$$\newcommand{\+}{^{\dagger}}% \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}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \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}} $$
$\endgroup$ $\begingroup$Similar to Felix Marin's solution but using generalized harmonic numbers and partial sums$$\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.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ $\begingroup$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!
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