Sum of Reciprocals of Squares plus 1

Theorem

 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1} = \frac 1 2 \left({\pi \coth \pi - 1}\right)$

Proof
Let $f \left({x}\right)$ be the real function defined on $\left({-\pi \,.\,.\, \pi}\right]$ as:


 * $f \left({x}\right) = e^x$

From Fourier Series: $e^x$ over $\left({-\pi \,.\,.\, \pi}\right)$, we have:
 * $\displaystyle f \left({x}\right) \sim \frac {\sinh \pi} \pi \left({1 + 2 \sum_{n \mathop = 1}^\infty \left({-1}\right)^n \left({\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} }\right)}\right)$

Setting $x = 0$, we have: