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What is the value of $$\sum_{n=1}^{\infty}\frac1{1+n^2}$$ I don't know how to solve it.
I tried to solve it by power series, but I failed.

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    $\begingroup$ I'm afraid it's one of those "inspection" answers, where you have to draw upon your knowledge of familiar and well known series formulas -- probably a disappointing solution! $\endgroup$ – Hugh Entwistle Oct 15 '16 at 11:07
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Hint: $$\frac{\pi x}{2}\coth \pi x=\frac{1}{2}+\sum_{n=1}^{\infty }\frac{1}{\left ( \frac{n}{x} \right )^2+1}$$ at $x=1$ $$\sum_{n=1}^{\infty }\frac{1}{n^2+1}=\frac{\pi}{2}\coth \pi -\frac{1}{2}$$

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I will give an elementary solution based on the following lemmas:

Lemma 1. For any $n\in\mathbb{N}^+$, $$ \int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-x}\,dx = \frac{1}{1+n^2}.$$ Proof: integration by parts.

Lemma 2. The following series is the Fourier series of a $2\pi$-periodic sawtooth wave $f(x)$,
that over the interval $(0,2\pi)$ equals $\frac{\pi-x}{2}$: $$ f(x)=\sum_{n\geq 1}\frac{\sin nx}{n} $$ Proof: direct computation of Fourier coefficients.

By Lemma 1 and Lemma 2, $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{n^2+1}&=&\int_{0}^{+\infty}f(x)e^{-x}\,dx\\(\text{periodicity of } f(x))\qquad&=&\left(1+\frac{1}{e^{2\pi}}+\frac{1}{e^{4\pi}}+\ldots\right)\int_{0}^{2\pi}f(x)e^{-x}\,dx\\[0.2cm]&=&\frac{e^{2\pi}}{e^{2\pi}-1}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx\\[0.2cm](\text{direct computation})\qquad &=&\color{red}{\frac{\pi\coth\pi-1}{2}}.\end{eqnarray*} $$

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Hint. It's well-known that $$\pi\cot(\pi z)=\frac{1}{z}+2\sum_{n=1}^\infty \frac{z}{z^2-n^2}$$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{1 \over n^{2} + 1} & = {1 \over 2\ic}\sum_{n = 0}^{\infty}\pars{{1 \over n + 1 - \ic} - {1 \over n + 1 + \ic}} \\[5mm] & = {1 \over 2\ic}\bracks{\vphantom{\Large A}\Psi\pars{1 + \ic} - \Psi\pars{1 - \ic}}\qquad \pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = {1 \over 2\ic}\braces{\vphantom{\LARGE A}% {1 \over \ic} - \bracks{\vphantom{\Large A}\Psi\pars{1 - \ic} - \Psi\pars{\ic}}}\qquad \pars{~\Psi\ Recurrence~} \\[5mm] & = -\,{1 \over 2} - {1 \over 2\ic}\, \bracks{\vphantom{\large A}\pi\cot\pars{\pi\ic}}\qquad \pars{~Euler\ Reflection\ Formula~} \\[5mm] & = -\,{1 \over 2} - {\pi \over 2\ic}\,\bracks{\vphantom{\large A}-\ic\coth\pars{\pi}} = \bbx{\ds{\pi\coth\pars{\pi} - 1 \over 2}} \end{align}

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  • $\begingroup$ This is exactly how I would do it :) $\endgroup$ – Arturo don Juan Oct 18 '16 at 0:56
  • $\begingroup$ @ArturodonJuan Thanks. It's a straightforward method and it's widely used ( and explained ) in the old Abramowitz and Stegun Table. $\endgroup$ – Felix Marin Oct 18 '16 at 0:58
  • $\begingroup$ Yeah I know, I'm a physics student who learned it from Arfken's book (and they cite Abramowitz and Stegun) $\endgroup$ – Arturo don Juan Oct 18 '16 at 1:03
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It is shown in this answer that $$ \mathrm{PV}\sum_{n\in\mathbb{Z}}\frac1{n+z}=\pi\cot(\pi z)\tag{1} $$ converges for all $z\in\mathbb{C}$. Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac1{n^2+z^2} &=\frac i{2z}\sum_{n=1}^\infty\left(\frac1{n+iz}+\frac1{-n+iz}\right)\\ &=\frac i{2z}\left[\mathrm{PV}\sum_{n\in\mathbb{Z}}\frac1{n+iz}-\frac1{iz}\right]\\ &=-\frac1{2z^2}+\frac{\pi i}{2z}\cot(\pi iz)\\[6pt] &=-\frac1{2z^2}+\frac\pi{2z}\coth(\pi z)\tag{2} \end{align} $$ Plugging in $z=1$ gives $$ \sum_{n=1}^\infty\frac1{n^2+1}=-\frac12+\frac\pi2\coth(\pi)\tag{3} $$

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