I have found that the fourier cosine series from $({-\pi},{\pi})$ of the function $f(x)=cosh(x)$ is

$$ \frac{2sinh({\pi})}{\pi}[1+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{n^2+1}cos(nx)]$$

How do I use this to show:

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

I have no idea really, the $coth{\pi}$ has thrown me off

I have found that the Fourier cosine series from $({-\pi},{\pi})$ of the function $f(x)=\cosh(x)$ is

$$ \frac{2\sinh({\pi})}{\pi}\left[\frac{1}{2}+ \sum_{n\: =\: 1}^{\infty}\:\ \frac{(-1)^n}{n^2+1}\cos(nx)\right]$$

How do I use this to show:

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

I have no idea really, the $\coth{\pi}$ has thrown me off.