Show ${\pi\over \sin(\pi z)}=\lim_\limits{m\to \infty} {\sum{{(-1)^n}\over z-n}}$ Showin: ${\pi\over \sin(\pi z)}=\lim_\limits{m\to \infty} {\sum_{m\ge |n|}{{(-1)^n}\over z-n}}$
A question before that said that $\sum_{n\in \Bbb{Z}}(-1)^nf(n)=-\sum_{a\text{ pole of f}}res_{a}({\pi\over \sin (\pi z)}f(z))$ for $f$ with no poles in $\Bbb{Z}$ but I really can't seem to see any relation as here there is no such $f$. Also, other questions regarding the above, pretty popular expression, was not relevant as they were not really about complex analysis with residues and singularities. I also don't quite understand why the limit could not be avoided by simply writing an infinite series, is it about convergence? Would appreciate any reference. 
 A: The reason of the confusion is that the letter $z$ has two different roles in the question. I will use $w$ for the complex variable $z$ for the parameter in the statement. So, from this point, $z$ is a fixed complex number that is not an integer.
Let $f(w)=\frac1{z-w}$. The function 
$\frac{\pi}{\sin(\pi w)}\cdot\frac1{z-w}$ has simple poles at the integers and at the point $z$, and we have
$$
\mathop{\mathrm{res}}\limits_{w=n} \left(\frac{\pi}{\sin(\pi w)}\cdot\frac1{z-w}\right) =
\frac{(-1)^n}{z-n}
\quad\text{for $n\in\mathbb{Z}$} \quad\text{and}\quad
\mathop{\mathrm{res}}\limits_{w=z} \left(\frac{\pi}{\sin(\pi w)}\cdot\frac1{z-w}\right) =
-\frac{\pi}{\sin(\pi z)}.
$$
Knowing that $\sin(w)\to\infty$ as $|\mathrm{Im}\,w|\to\infty$,
you can apply the residue theorem in the circle $|w|=m+\frac12$ with some ``big'' positive integer (or a rectangle with vertices $\pm(m+\frac12)\pm\sqrt{m}\,i$ etc.) to see that the sum of residues is
$$
\lim_{m\to \infty}\left(\sum_{|n|\le m}\frac{(-1)^n}{z-n} -\frac{\pi}{\sin(\pi z)}\right)=0.
$$
Concerning the second question, usually we like absolute convergent series better, but you are right, in this particular case we could omit taking limits. On the other hand, the limit version of the question may give you a hint: apply the residue theorem in the circle $|z|=m+\frac12$.

I prefer a different way to prove the formula. Briefly, let
$$
g(z) = \frac{\pi}{\sin(\pi z)} - 
\lim_{m\to \infty}\sum_{|n|\le m}\frac{(-1)^n}{z-n}.
$$
It is easy to see that $g$ has only removable singularities, it is periodic and bounded, so due to Liouville's theorem, it must be constant; but $g$ is odd, therefore $g\equiv0$.
A: Is a particular case of the Mittag-Leffler's theorem. See How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$? for a similar example.
