This is a homework question whereby I am supposed to evaluate:

$$\sum_{n=1}^\infty \frac{1}{n^2 +1}$$

Wolfram Alpha outputs the answer as

$$\frac{1}{2}(\pi \coth(\pi) - 1)$$

But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.

  • $\begingroup$ Have you studied the method of series calculation using Fourier series? $\endgroup$
    – Frank
    Apr 2, 2014 at 16:45
  • $\begingroup$ (no idea how to use fourier series)You can estimate through $\arctan(x)$ though $\endgroup$
    – Guy
    Apr 2, 2014 at 16:46
  • 1
    $\begingroup$ @Frank What's the function that works here? $\endgroup$
    – Git Gud
    Apr 2, 2014 at 16:47
  • 1
    $\begingroup$ @Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious. $\endgroup$
    – Briandmg
    Apr 2, 2014 at 16:48
  • 11
    $\begingroup$ This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^{-1}\sin x=\prod_k \Big(1-x^2(\pi k)^{-2}\Big)\Rightarrow (i\pi x)^{-1}\sin i\pi x=\prod_k \Big(1+x^2k^{-2}\Big)$$ Take logs, differentiate, and tada: $\sum (k^2+x^2)^{-1}=\frac{1}{2x^2}(\pi x\coth \pi x-1)$ $\endgroup$
    – jcg
    Apr 2, 2014 at 16:51

5 Answers 5


Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum

We can solve a more general sum, $$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = \frac{\pi}{a} \coth(\pi a).$$

Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so $$\sum_{n=-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\pi\left[\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) + \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right)\right].$$ Computing the residues: $$\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) = \lim_{z\rightarrow ia}\frac{(z-ia)\cot(\pi z)}{(z-ia)(z+ia)} = \frac{\cot(\pi ia)}{2i a} $$ and $$ \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right) = \lim_{z\rightarrow -ia}\frac{(z+ia)\cot(\pi z)}{(z+ia)(z-ia)} = \frac{\cot(i\pi a)}{2ia}.$$ Therefore, summing these we get $$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\frac{\pi\cot(i\pi a)}{ia} = \frac{\pi \coth(\pi a)}{a}.$$

You should be able to extend this idea to your sum with some effort.

  • 1
    $\begingroup$ The following MSE link describes the same method if the OP is interested in seeing another example. $\endgroup$ Apr 2, 2014 at 19:44

We can start from the Weierstrass product for the $\sinh$ function: $$\frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1} $$ then consider the logarithmic derivative of both sides. This leads to: $$\coth z-\frac{1}{z}=\sum_{n=1}^{+\infty}\frac{2z}{z^2+\pi^2 n^2}\tag{2} $$ or to: $$\pi\coth(\pi w)-\frac{1}{w}=\sum_{n=1}^{+\infty}\frac{2w}{w^2+ n^2}.\tag{3} $$ Now just set $w=1$ in $(3)$.

Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $\sum_{n\geq 0}\frac{1}{n^2+1}$ is simply related to $\sum_{n\geq 0}e^{-\pi n}$, which is a geometric series.

  • $\begingroup$ This is so beautiful. Thank you. $\endgroup$ Mar 17, 2021 at 20:16

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \sum_{n = 1}^{\infty}{1 \over n^{2} + 1} & = \sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + \ic}\pars{n + 1 - \ic}} ={\Psi\pars{1 + \ic} - \Psi\pars{1 - \ic} \over \pars{1 + \ic} - \pars{1 - \ic}} \\[5mm] & =\Im\Psi\pars{1 + \ic} \end{align} where $\Psi\pars{z}$ is the Digamma Function.

With the identity $\ds{\Im\Psi\pars{1 + \ic y} = -\,{1 \over 2y} + \half\,\pi\coth\pars{\pi y}}$ we'll have: $$\color{#00f}{\large% \sum_{n = 1}^{\infty}{1 \over n^{2} + 1} = \half\bracks{\pi\coth\pars{\pi} - 1}} $$


There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:


$$ f(z) = \frac {\pi} {(1+z^2)\tan(\pi z)} $$

Then $f$ has simple poles $\forall n \in \mathbb{Z}$ and also at $\pm i$.

You can calculate the residues as

$$ \text{Res}(f(z), n\pi) = \frac {1} {1 + n^2} $$


$$ \text{Res}(f(z), \pm i) = \frac {-\pi} {2tanh(\pi)} $$

If these are hard to calculate for you, I can give you more detail.

Now, let $\Gamma _N$ be the square contour with vertices $(N + \frac 1 2) (\pm 1 \pm i)$

Then Cauchy's Residue Theorem tells us

$$ \int_{\Gamma N} f(z) dz = 2\pi i \sum \text{Res}(f(z), z) $$

Where the sum is across all the poles inside the contour.

Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $\pm i$.


$$ \int_{\Gamma N} f(z) dz = 2\pi i \left [ \frac {-2\pi} {2 \tanh(\pi)} + \sum_{n = -N}^{N} \frac{1} {1 + n^2} \right ] $$

Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $\frac {\pi} {tan(\pi z)}$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.

Then by the Estimation Theorem we would have that

$$ \left | \int_{\Gamma_N} f(z) dz \right | \leq \text{length}(\Gamma_N) \text{sup}_{z \in \Gamma _N} |f(z)| \leq 4(2N + 1) C \text{sup}_{z \in \Gamma_N} \left\| \frac {1} {1 + z^2}\right \| \leq \frac {4C(2N + 1)} { 1 + N^2} = O(\frac 1 N) $$

So we let $N \to \infty$ then we get

$$ 0 = 2\pi i \left [ \frac {-2\pi} {2 \tanh(\pi)} + \sum_{n = -\infty}^{\infty} \frac{1} {1 + n^2} \right ] $$ So then we have $$ 0 = \frac {-\pi} { \tanh(\pi)} + 2\sum_{n = 1}^{\infty} \frac{1} {1 + n^2} + 1 $$ Where the 1 that has randomly appeared is the $n = 0 $ term

So $$ \sum_{n = 1}^{\infty} \frac{1} {1 + n^2} = \frac 1 2 \left [\frac {\pi} { \tanh(\pi)} - 1 \right] = \frac 1 2 (\pi \coth (\pi) - 1). $$

I hope you know some complex analysis otherwise this might have meant nothing to you...

This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.

If you want to evaluate

$$ \sum_{n = 1} ^ {\infty} \phi(n) $$

Where \phi can easily be extended to all of $\mathbb {C}$ you just take $$f(z) = \frac {\pi} {\phi(z) \tan(\pi z)} $$ And do the same thing, and if you want to evaluate $$\sum_{n = 1} ^ {\infty} (-1)^n \phi(n) $$ You just take $$f(z) = \frac {\pi} {\phi(z) \sin(\pi z)} $$ In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.


This is related to this answer, where it is shown that $$ \sum_{k=1}^\infty\frac{1}{k^2-z^2} =\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]\tag{1} $$ Equation $(1)$ is valid for all $z\in\mathbb{C}$, in particular for $z=i$, which gives $$ \sum_{k=1}^\infty\frac{1}{k^2+1} =\frac12\left[\pi\coth(\pi)-1\right]\tag{2} $$


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