Show that $\frac{\pi}{\sin\pi z}=\lim_{m\to\infty}\sum_{n=-m}^m(-1)^n\frac{1}{z-n}$ Question: Show that $\frac{\pi}{\sin\pi z}=\lim_{m\to\infty}\sum_{n=-m}^m(-1)^n\frac{1}{z-n}$.
Specifically.....: Ahlfors derives this on page 189/190 of his "Complex Analysis" book.  In it, he uses his derivations for $\pi\cot\pi z$, specifically, $$\pi\cot\pi z=\lim_{m\rightarrow\infty}\sum_{n=-m}^m\frac{1}{z-n}=\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2}$$  I suppose I am wondering if there is maybe a way to derive the question above without the use of $\pi\cot\pi z$?  Maybe using the Laurent Series expansion of $\frac{\pi}{\sin\pi z}$, but I wasn't getting anything to come out cleanly (maybe I just made a mistake).
 A: As already noted in the comments this is a typical usage of the Mittag-Leffler theorem. As the same theorem also can be used for deriving the cited cotangent series this is not a fundamentally different approach.
Note, however, that the cotangent series can be derived completely different using the so-called Herglotz Trick, which is an elementary approach using no heavy machinery from complex analysis. The linked PDF only proves the series representation in case of real arguments but if I am not mistaken the result can be easily extended to whole complex plane by continuation as usual. This might be of particular interest if you have not covered Mittag-Leffler yet.
The Laurent series approach fails due to the many singularities of the given function (as noted by Ninad Munshi). I am not aware of a different approach than using Mittag-Leffler directly or the Herglotz Trick.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{{\pi \over \sin\pars{\pi z}} =
\lim_{m \to \infty}\sum_{n = -m}^{m}{\pars{-1}^{n} \over z - n}}:\
{\Large ?}}$.

With $\ds{m \in \mathbb{N}_{\ \geq\ 1}}$:
\begin{align}
&\bbox[5px,#ffd]{\pi \over \sin\pars{\pi z}} =
{1 \over z} + {1 \over z}\bracks{{\pi z \over \sin\pars{\pi z}} - 1}
\\[5mm] = &\
{1 \over z} + {1 \over z}\bracks{%
\lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m}
a_{n}\pars{{1 \over z - n} + {1 \over n}}}
\end{align}
where $\ds{a_{n} \equiv
\mrm{Res}\pars{{\pi z \over \sin\pars{\pi z}} - 1, z = n} =
\lim_{z \to n}
\braces{\pars{z - n}\bracks{{\pi z \over \sin\pars{\pi z}} - 1}} = \pars{-1}^{n}\, n}$.
See Mittag-Leffler Expansion.
Then,
\begin{align}
&\bbox[5px,#ffd]{\pi \over \sin\pars{\pi z}} =
{1 \over z} + {1 \over z}\bracks{%
\lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m}
{\pars{-1}^{n}\,z \over z - n}} =
{1 \over z} + 
\lim_{m \to \ \infty}\sum_{{\Large n\ =\ -m} \atop {\Large n\ \not=\ 0}}^{m}
{\pars{-1}^{n} \over z - n}
\\[5mm] = &\
\bbx{\large\lim_{m \to \ \infty}\sum_{n\ =\ -m}^{m}
{\pars{-1}^{n} \over z - n}} \\ &
\end{align}
