This is a homework question whereby I am supposed to evaluate:
$$\sum_{n=1}^\infty \frac{1}{n^2 +1}$$
Wolfram Alpha outputs the answer as
$$\frac{1}{2}(\pi \coth(\pi) – 1)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
Best Answer
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.