Sum to closed form I need to evaluate the following summation:
$$
\sum_{n\in\mathbb{Z}} \frac{-1}{i(2n+1)\pi -\mu}
$$
where $n$ is summed over all the integers from $-\infty$ to $\infty$ including 0. Putting this into Mathematica gives $\frac{1}{2}\tanh\frac{\mu}{2}$. What is the intermediate steps to get from the summation to the closed form?
 A: $$\sum_{n\in\mathbb{Z}} \frac{-1}{i(2n+1)\pi -\mu}(\frac{i(2n+1)\pi+\mu}{i(2n+1)\pi+\mu})=\sum_{n\in\mathbb{Z}} \frac{i(2n+1)\pi+\mu}{(2n+1)^2\pi^2 +\mu^2}$$
$$=\sum_{n\in\mathbb{Z}} \frac{i(2n+1)\pi}{(2n+1)^2\pi^2 +\mu^2}+\frac{\mu}{(2n+1)^2\pi^2 +\mu^2}$$
the first term will take the following values
$$\sum_{n\in\mathbb{Z}} \frac{i(2n+1)\pi}{(2n+1)^2\pi^2+\mu^2}=\frac{\pi*i}{\pi^2+\mu^2}+\frac{3\pi*i}{3^2\pi^2+\mu^2}+...-\frac{\pi*i}{\pi^2+\mu^2}-\frac{3\pi*i}{3^2\pi^2+\mu^2}-...=0$$
the second terms equal to
$$\sum_{n\in\mathbb{Z}} \frac{\mu}{(2n+1)^2\pi^2 +\mu^2}=2\sum_{n=0}^{\infty }\frac{\mu}{(2n+1)^2\pi^2 +\mu^2}=2(\frac{1}{4}\tanh\frac{\mu}{2})=\frac{1}{2}\tanh\frac{\mu}{2}$$
A: Start with Euler's infinite product formula for the sine function:
$$\text{sin }z=z \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \pi^2}\right).$$
It follows that
$$\text{sinh}\;z = -i\;\text{sin}\;iz=z \prod_{n=1}^\infty \left(1+\frac{z^2}{n^2 \pi^2}\right).$$
As a result,
$$\text{sinh}\;\frac{z}{2} = \frac{z}{2} \prod_{n=1}^\infty \left(1+\frac{z^2}{4n^2 \pi^2}\right) = \frac{z}{2} \prod_{n\;\text{even}\;\geq 2} \left(1+\frac{z^2}{n^2 \pi^2}\right).$$
Using the identity $$\text{sinh}\;2z=2\;\text{sinh}\;z\;\text{cosh}\;z,$$
we find that
$$\text{cosh}\;\frac{z}{2} = \frac{\text{sinh}\;z}{2\;\text{sinh}\;\frac{z}{2}}=\frac{z \prod_{n=1}^\infty \left(1+\frac{z^2}{n^2 \pi^2}\right)}{2\cdot\frac{z}{2} \prod_{n\;\text{even}\;\geq 2} \left(1+\frac{z^2}{n^2 \pi^2}\right)}=\prod_{n\;\text{odd}\;\geq 1} \left(1+\frac{z^2}{n^2 \pi^2}\right).$$
Taking the logarithmic derivative $\frac{f'}{f}$ on both sides yields
$$\frac{1}{2}\text{tanh}\;\frac{z}{2}=\sum_{n\;\text{odd}\;\ge 1}\frac{2z/(n^2 \pi^2)}{1+\frac{z^2}{n^2 \pi^2}}=\sum_{n\;\text{odd}\;\ge 1}\frac{2z}{n^2 \pi^2+z^2}.$$
Now,
$$\frac{2z}{n^2 \pi^2+z^2}=\frac{1}{z+n \pi i}+\frac{1}{z-n \pi i},$$
so
$$\frac{1}{2}\text{tanh}\;\frac{z}{2}=\sum_{n\;\text{odd}\;\ge 1}\left(\frac{1}{z+n \pi i}+\frac{1}{z-n \pi i}\right),$$
which is the same as the desired sum.
(There are convergence issues with the doubly-infinite sum in the question as stated, but this is the principal value of the sum, which is probably what was intended.)
