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 $g\equiv0$.
