Question: 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.

Answer: 
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**,
$$\begin{eqnarray\*}\sum\_{n\geq 1}\frac{1}{n^2+1}&=&\int\_{0}^{+\infty}f(x)e^{-x}\,dx\\(\text{periodicity of } f(x))\qquad&=&\left(1+\frac{1}{e^{2\pi}}+\frac{1}{e^{4\pi}}+\ldots\right)\int\_{0}^{2\pi}f(x)e^{-x}\,dx\\[0.2cm]&=&\frac{e^{2\pi}}{e^{2\pi}-1}\int\_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx\\[0.2cm](\text{direct computation})\qquad &=&\color{red}{\frac{\pi\coth\pi-1}{2}}.\end{eqnarray\*} $$
It is shown in [this answer](https://math.stackexchange.com/a/143179) 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
$$
\sum\_{n=1}^\infty\frac1{n^2+1}=-\frac12+\frac\pi2\coth(\pi)\tag{3}
$$