Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$

Solution 1:

Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum

We can solve a more general sum, $$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = \frac{\pi}{a} \coth(\pi a).$$

Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so $$\sum_{n=-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\pi\left[\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) + \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right)\right].$$ Computing the residues: $$\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) = \lim_{z\rightarrow ia}\frac{(z-ia)\cot(\pi z)}{(z-ia)(z+ia)} = \frac{\cot(\pi ia)}{2i a} $$ and $$ \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right) = \lim_{z\rightarrow -ia}\frac{(z+ia)\cot(\pi z)}{(z+ia)(z-ia)} = \frac{\cot(i\pi a)}{2ia}.$$ Therefore, summing these we get $$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\frac{\pi\cot(i\pi a)}{ia} = \frac{\pi \coth(\pi a)}{a}.$$

You should be able to extend this idea to your sum with some effort.

Solution 2:

We can start from the Weierstrass product for the $\sinh$ function: $$\frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right)\tag{1} $$ then consider the logarithmic derivative of both sides. This leads to: $$\coth z-\frac{1}{z}=\sum_{n=1}^{+\infty}\frac{2z}{z^2+\pi^2 n^2}\tag{2} $$ or to: $$\pi\coth(\pi w)-\frac{1}{w}=\sum_{n=1}^{+\infty}\frac{2w}{w^2+ n^2}.\tag{3} $$ Now just set $w=1$ in $(3)$.

Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $\sum_{n\geq 0}\frac{1}{n^2+1}$ is simply related to $\sum_{n\geq 0}e^{-\pi n}$, which is a geometric series.