How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \dfrac{1}{n^2}=\frac{\pi^2}{6}$$
But I'm stuck at this problem.
First consider the following expansion of $\pi \cot(\pi z)$:
$$\pi \cot(\pi z) = \frac{1}{z} + \sum_{n = 1}^\infty \frac{2z}{z^2 - n^2} \quad (z \neq 0, \pm 1, \pm 2,\ldots)$$
Replacing $z$ by $iz$, we have
$$-i\pi \coth(\pi z) = \frac{1}{iz} - \sum_{n = 1}^\infty \frac{2iz}{z^2 + n^2} = -i\left(\frac{1}{z} + 2\sum_{n = 1}^\infty \frac{z}{z^2 + n^2}\right)$$
Thus
$$\pi \coth(\pi z) = \frac{1}{z} + 2\sum_{n = 1}^\infty \frac{z}{z^2 + n^2} \quad (z \neq 0, \pm i, \pm 2i,\ldots)$$
Evaluting at $z = 1$ results in
$$\pi \coth(\pi) = 1 + 2\sum_{n = 1}^\infty \frac{1}{1 + n^2},$$
or
$$\sum_{n = 1}^\infty \frac{1}{1 + n^2} = \frac{\pi \coth(\pi) - 1}{2}.$$
Therefore
\begin{align}\sum_{n = 0}^\infty \frac{1}{1 + n^2} &= \frac{\pi\coth(\pi)+ 1}{2}\\ &= \frac{1}{2}\left(\frac{\pi(e^{\pi} + e^{-\pi})}{e^{\pi} - e^{-\pi}} + 1\right)\\ &= \frac{1}{2}\left(\frac{\pi(e^{2\pi} + 1)}{e^{2\pi} - 1} + 1\right)\\ &= \frac{(\pi + 1)e^{2\pi} + (\pi - 1)}{2(e^{2\pi} - 1)}\\ &= \frac{(\pi + 1)(e^{2\pi} - 1) + 2\pi}{2(e^{2\pi} - 1)}\\ &= \frac{\pi + 1}{2} + \frac{\pi}{e^{2\pi} - 1}.\\ \end{align}
Hint: Differentiate the natural logarithm of Euler's infinite product expression for the sine function, then use the well-known relations between trigonometric and hyperbolic functions.
