What is the value of the series $\sum_{n\geq 1}\frac{1}{n^2+1}$? 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.
 A: 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,
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{n^2+1}&=&\int_{0}^{+\infty}f(x)e^{-x}\,dx\\(\text{periodicity of } f(x))\qquad&=&\left(1+\frac{1}{e^{2\pi}}+\frac{1}{e^{4\pi}}+\ldots\right)\int_{0}^{2\pi}f(x)e^{-x}\,dx\\[0.2cm]&=&\frac{e^{2\pi}}{e^{2\pi}-1}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx\\[0.2cm](\text{direct computation})\qquad &=&\color{red}{\frac{\pi\coth\pi-1}{2}}.\end{eqnarray*} $$
A: 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 $x=1$
$$\sum_{n=1}^{\infty }\frac{1}{n^2+1}=\frac{\pi}{2}\coth \pi -\frac{1}{2}$$
A: Hint. It's well-known that
$$\pi\cot(\pi z)=\frac{1}{z}+2\sum_{n=1}^\infty \frac{z}{z^2-n^2}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\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}
A: 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
$$
\sum_{n=1}^\infty\frac1{n^2+1}=-\frac12+\frac\pi2\coth(\pi)\tag{3}
$$
