Value of $\sum_{k=1}^{\infty}\frac{1}{k^2+a^2}$ So my question is to find the value of $\sum_{k=0}^\infty\frac{1}{k^2+1}$ and more generally $\sum_{k=0}^\infty\frac{1}{Q(k)}$ 
where Q is a quadratic polynomial with no zeroes on the integers.
I can prove it converges, by the comparison test. I think I 've seen such a sum before, and I think it has a hyperbolic function solution. But the way they did it there involved complex analysis, and I'm not very comfortable with that.

Using the residue theorem, you can show that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \coth{\pi a}$$
This is equivalent to saying that
$$\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} = \frac{1}{2} \left (\frac{\pi}{a} \coth{\pi a} - \frac{1}{a^2}\right )$$
You can also derive this by considering the Maclurin expansion of $z \coth{z}$:
$$z \coth{z} = 1 + \sum_{k=1}^{\infty} \frac{B_{2 k} (2 z)^{2 k}}{(2 k)!}$$
where $B_{2 k}$ is a Bernoulli number, which also shows up in Riemann zeta functions of even, positive argument:
$$\zeta(2 k) = (-1)^{k+1} \frac{B_{2 k} (2 \pi)^{2 k}}{2 (2 k)!}$$
To evaluate the sum, factor out $n^2$ from the denominator and Taylor expand:
$$\begin{align}\sum_{n=1}^{\infty} \frac{1}{n^2+a^2} &= \frac{1}{n^2} \frac{1}{1+ \frac{a^2}{n^2}}\\ &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} (-1)^k \left (\frac{a^2}{n^2}\right )^{k} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k}\sum_{n=1}^{\infty} \frac{1}{n^{2 k+2}} \\ &=\sum_{k=0}^{\infty} (-1)^k a^{2 k} \zeta(2 k+2)\\ &= \frac{1}{2 a^2}\sum_{k=1}^{\infty} \frac{B_{2 k} (2 \pi a)^{2 k}}{(2 k)!} \\ &= \frac{1}{2 a^2} ( \pi a \coth{\pi a} - 1)\\  \end{align}$$
The result follows.
