Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$
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.
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 with $\sum_{n\geq 0}e^{-\pi n}$, which is a geometric series.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \sum_{n = 1}^{\infty}{1 \over n^{2} + 1} & = \sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + \ic}\pars{n + 1 - \ic}} ={\Psi\pars{1 + \ic} - \Psi\pars{1 - \ic} \over \pars{1 + \ic} - \pars{1 - \ic}} \\[5mm] & =\Im\Psi\pars{1 + \ic} \end{align} where $\Psi\pars{z}$ is the Digamma Function.
With the identity $\ds{\Im\Psi\pars{1 + \ic y} = -\,{1 \over 2y} + \half\,\pi\coth\pars{\pi y}}$ we'll have: $$\color{#00f}{\large% \sum_{n = 1}^{\infty}{1 \over n^{2} + 1} = \half\bracks{\pi\coth\pars{\pi} - 1}} $$