Evaluate the series - 2

The Fourier series of $e^x$ at $x=\pm \pi$ is $\frac{1}{\pi}\sinh\pi+\frac{2\sinh\pi}{\pi}\sum_{n=1}^{\infty}\frac{1}{1+n^2}=\frac{e^{\pi}+e^{-\pi}}{2}$ Solving for $\sum_{n=0}^{\infty}\frac{1}{1+n^2}$ gives $\frac{1}{2}(1+\pi \coth\pi) \approx \boxed{2.07667}$

Doing it the long way (also the only way I know) using digamma function .

$\begin{aligned} S & = \sum_{n=0}^\infty \frac 1{1+n^2} = \sum_{n=0}^\infty \frac 1{(n-i)(n+i)} \\ & = \frac i2 \sum_{\color{#3D99F6}n=0}^\infty \left(\frac 1{n+i} - \frac 1{n-i} \right) & \small \color{#3D99F6} \text{Digamma function } \psi (z+1) = - \gamma + \sum_{k=1}^\infty \left(\frac 1k - \frac 1{k+z} \right) \\ & = 1 + \frac i2 \sum_{\color{#D61F06}n=1}^\infty \left(\frac 1{n+i} - \frac 1{n-i} \right) & \small \color{#3D99F6} \text{where } \gamma \text{ is the Euler-Mascheroni constant.} \\ & = 1 + \frac i2 (\psi(1-i) -\psi (1+i)) & \small \color{#3D99F6} \text{Using } \psi (1-z) - \psi (z) = \pi \cot (\pi z) \\ & = 1 + \frac i2 \left(\pi \cot (\pi i) - \frac 1i\right) & \small \color{#3D99F6} \text{and } \psi (1+z) = \psi (z) + \frac 1z \\ & = 1 + \frac \pi 2 \coth \pi - \frac 12 \\ & \approx \boxed{2.077} \end{aligned}$