So I have been trying for a few days to figure out the sum of

$$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical analysis and CAS that this sum equals

$$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$

But I have not been able to prove this yet. Anyone know how? My guess is that the sum of this series is related to fourier-series but nothing particalr comes to mind.

For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.

  • 11
    $\begingroup$ There is the identity $$\frac{\sin\,\pi x}{\pi x}=\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)$$ Take the logarithm, differentiate, see what happens. $\endgroup$ May 5, 2012 at 19:17
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    $\begingroup$ Why was this downvoted? $\endgroup$
    – t.b.
    May 5, 2012 at 19:22
  • $\begingroup$ Did you mean $k=1$ underneath the sum sign? $\endgroup$ May 5, 2012 at 19:30
  • $\begingroup$ Yes he did, @Joel. $\endgroup$ May 5, 2012 at 19:33
  • $\begingroup$ @N3buchadnezzar your expected answer has an error. Checked against Wolfram Alpha's snipurl.com/23dq2hp and instead of plus sign, it shows minus sign. Follow J.M's hint, your might approach the answer. $\endgroup$ May 5, 2012 at 20:19

5 Answers 5


This question was settled in the Mathematics chatroom, but I'll put up the solution here for reference.

Starting with the infinite product

$$\frac{\sin\,\pi x}{\pi x}=\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)$$

taking the logarithm of both sides gives

$$\log\left(\frac{\sin\,\pi x}{\pi x}\right)=\log\left(\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2}\right)\right)=\sum_{k=1}^\infty \log\left(1-\frac{x^2}{k^2}\right)$$

Differentiation gives

$$\frac{\pi x}{\sin\,\pi x}\left(\frac{\cos\,\pi x}{x}-\frac{\sin\,\pi x}{\pi x^2}\right)=\sum_{k=1}^\infty \frac{-2x}{k^2\left(1-\frac{x^2}{k^2}\right)}$$

which simplifies to

$$\pi\cot\,\pi x-\frac1{x}=-2x\sum_{k=1}^\infty \frac1{k^2-x^2}$$


$$\sum_{k=1}^\infty \frac1{k^2-x^2}=\frac1{2x^2}-\frac{\pi\cot\,\pi x}{2x}$$

  • $\begingroup$ While we are taking logarithm should'nt one be careful? i.e $\frac{\sin(\pi x)}{\pi x}$ can be negative real number for example. What branch are we considering? $\endgroup$ May 1, 2020 at 13:50
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    $\begingroup$ @S.S "can be negative real number" - have you looked at a plot of $\dfrac{\sin\pi x}{\pi x}$ for $0<x<1$ (cf. the title of this question)? $\endgroup$ May 1, 2020 at 14:43
  • $\begingroup$ Sorry I forgot that that the domain you were talking about was $x\in(0,1)$. $\endgroup$ May 3, 2020 at 12:05

Let us consider the principal value of the conditionally convergent infinite harmonic series $$ \begin{align} f(z) &=\sum_{k=-\infty}^\infty\frac{1}{z+k}\\ &=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k}\tag{1a}\\ &=\lim_{n\to\infty}\frac1z+\sum_{k=1}^n\frac{1}{z-k}+\frac{1}{z+k}\tag{1b}\\ &=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1c} \end{align} $$ The series in $(1c)$ converges absolutely for all non-integer $z$.

Each of the terms in $(1c)$ is odd, so $f(-z)=-f(z)$.

The series in $(1a)$ shows that $f$ has a simple pole with residue $1$ at each integer.

$f$ has period $1$: $$ \begin{align} f(z+1)-f(z) &=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{z+k+1}-\frac{1}{z+k}\\ &=\lim_{n\to\infty}\frac{1}{z+n+1}-\frac{1}{z-n}\\ &=0\tag{2} \end{align} $$ $f(1/2)=0$: $$ \begin{align} f(1/2) &=\lim_{n\to\infty}\sum_{k=-n}^n\frac{1}{k+1/2}\\ &=\lim_{n\to\infty}\frac{1}{n+1/2}\\ &=0\tag{3} \end{align} $$ Take the derivative of $f$: $$ f'(z)=\lim_{n\to\infty}\sum_{k=-n}^n\frac{-1}{(z+k)^2}\tag{4} $$ This series converges absolutely. and the terms monotonically go to $0$ as $|\Im(z)|\to\infty$.

Let's consider $if(iy)$ as $y\to\infty$. Using $(1c)$, we get $$ \begin{align} if(iy) &=\frac1y+\sum_{k=1}^\infty\frac{2y}{y^2+k^2}\\ &=\frac1y+\sum_{k=1}^\infty\frac{2/y}{1+(k/y)^2}\tag{5} \end{align} $$ As $y\to\infty$, the summation in $(5)$ is a Riemann sum for the integral $$ \int_0^\infty\frac{2\mathrm{d}x}{1+x^2}=\pi\tag{6} $$ Thus, $if(iy)\to\pi$ as $y\to\infty$ and $if(iy)\to-\pi$ as $y\to-\infty$.

Since $f$ has period $1$ and $f'(z)\to0$ as $|\Im(z)|\to\infty$, it is evident that $f(z)\to-i\pi$ as $\Im(z)\to\infty$ and $f(z)\to i\pi$ as $\Im(z)\to-\infty$. This means that $f$ is bounded when away from the real axis.

The functions $f$ and $\pi\cot(\pi z)$ have the same poles, with identical residues, and both are bounded when away from the real axis. Thus, their difference is bounded for all $z$. Since their difference is analytic and bounded, it must be constant. This difference is $0$ at $1/2$, so it must be $0$ everywhere. Therefore, the principal value of $$ \sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)\tag{7} $$ for all $z$.

Combining $(1c)$ and $(7)$ yields $$ \frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}=\pi\cot(\pi z)\tag{8} $$ Therefore, $$ \begin{align} \sum_{k=1}^\infty\frac{1}{k^2-z^2} &=\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z}\\ &=\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]\tag{9} \end{align} $$

  • 1
    $\begingroup$ A real approach is given in this answer, $\endgroup$
    – robjohn
    Feb 23, 2021 at 14:19

You may prove this by expanding $\cos(zx)$ in Fourier series as shown here.

This paper could help too as well as articles in SE dealing with evaluation of $\zeta(n)$ with $n$ even.


@Kobe found here by means of residue calculus

$$\sum_{k = 1}^\infty \frac{1}{k^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2}$$

To get your result, replace $a$ by $ia$.


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

\begin{align} S & \equiv \bbox[5px,#ffd]{\sum_{k = 1}^{\infty} {1 \over k^{2} - a^{2}}} = \sum_{k = 0}^{\infty} {1 \over \pars{k + 1 - a}\pars{k + 1 + a}} \\[5mm] & = {\Psi\pars{1 - a} - \Psi\pars{1 + a} \over -2a} = {\Psi\pars{1 - a} - \bracks{\Psi\pars{a} + 1/a} \over -2a} \\[5mm] & = {1 \over -2a}\bracks{\pi\cot\pars{\pi a} - {1 \over a}} = \bbx{{1 \over 2a}\bracks{{1 \over a} - \pi\cot\pars{\pi a}}} \\ & \end{align}

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