Q:

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
$$\pi\cot(\pi z)=\frac{1}{z}+2\sum_{n=1}^\infty \frac{z}{z^2-n^2}$$

