<s>[INST] How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$? Does anyone know the general strategy for summing a series of the form:
$$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$
where $a$ is a positive integer?
Any hints or ideas would be great!
 [/INST] Certain series of the form $\sum_{n = -\infty}^\infty f(n)$ can be evaluated by means of residue calculus. One important result which is useful to this problem states that if $f$ is holomorphic on $\Bbb C \setminus \{z_1,\ldots, z_k\}$ (where the $z_i$ are the isolated singularities of $f$) and $|zf(z)|$ is bounded for $|z|$ sufficiently large, then $\sum_{n = -\infty}^\infty f(n)$ is the negative of the sum of the residues of $\pi \cot(\pi z)f(z)$ at the $z_i$, provided none of the $z_i$ are integers. You can find more information in Marsden's Basic Complex Analysis text.
To apply the result to this problem, first note that $$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{1}{2}\sum_{n = -\infty}^\infty \frac{1}{n^2 + a^2} - \frac{1}{2a^2}$$ Let $f(z) = \frac{1}{z^2 + a^2}$. Then $f$ has simple poles at $z = ai$ and $z = -ai$. For $|z| \ge \max\{1, 2a\}$, $$|zf(z)| \le \frac{|z|}{|z|^2 - a^2} \le \dfrac{|z|}{|z|^2 - \frac{|z|^2}{4}} = \frac{4}{3|z|} \le \frac{4}{3}$$
Therefore 
\begin{align}-\sum_{n = -\infty}^\infty \frac{1}{n^2 + a^2} &= \text{Res}_{z = ai} \frac{\pi\cot(\pi z)}{z^2 + a^2} + \text{Res}_{z = -ai} \frac{\pi\cot(\pi z)}{z^2 + a^2}\\
&= \frac{\pi\cot(\pi ai)}{2ai} - \frac{\pi\cot(-\pi ai)}{2ai}\\
&= -\frac{\pi\coth(\pi a)}{2a} - \frac{\pi \coth(\pi a)}{2a}\\
&= -\frac{\pi\coth(\pi a)}{a}\end{align} 
Hence,
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2}$$
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