Mittag Leffler partial fraction expansion for $\frac{\sin(z)}{\sin(\pi z)}$ I want to prove the following identity:
$$\dfrac{\sin(z)}{\sin(\pi z)} = \dfrac{1}{\pi} + \dfrac{z}{\pi}\sum_{n \neq 0}(-1)^n \dfrac{\sin(n)}{n(z-n)} $$
using Mittag-Leffler. 
I'm able to show that we have 
$$\dfrac{\sin(z)}{\sin(\pi z)} = \dfrac{z}{\pi}\sum_{n \neq 0}(-1)^n \dfrac{\sin(n)}{n(z-n)} +g(z)$$
 whee $g(z)$ is entire, so the goal would be to show that $g$ is bounded, hence constant. I don't see any periodicity conditions here that would help me. 
what kind of observations should I make here?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Simple poles of $\ds{\sin\pars{z}/\sin\pars{\pi z}}$ are at
  $\ds{n \in \mathbb{Z}\setminus\braces{0}}$ with correspondent residues
  $\ds{r_{n} = \lim_{z \to n}\bracks{\pars{z - n}\sin\pars{z}/\sin\pars{\pi z}} =
\pars{-1}^{n}\,\sin\pars{n}/\pi}$. Moreover,
  $\ds{\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty}{r_{n}/n^{2}}}$ is convergent

such that its Mittag-Leffler Expansion becomes
\begin{align}
{\sin\pars{z} \over \sin\pars{\pi z}} - {1 \over \pi} & =
\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty}
r_{n}\pars{{1 \over z - n} + {1 \over n}}
\\[5mm]
{\sin\pars{z} \over \sin\pars{\pi z}} & =
{1 \over \pi} + {1 \over \pi}
\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty}
\pars{-1}^{n}\,{\sin\pars{n} \over z - n}\ +\
\underbrace{{1 \over \pi}\sum_{{\large n = -\infty} \atop {\large n \not= 0}}^{\infty}
\pars{-1}^{n}\,{\mrm{sinc}\pars{n}}}_{\ds{\equiv \,\mrm{g}\pars{z} = \color{red}{\large 0}}}
\end{align}
A: Look at the limits as $\Im(z) \to \pm \infty$. Since $\pi>1$, the left-hand side tends to zero. If you can show that the sum on the right-hand side is bounded as $\Im(z) \to \pm \infty$, then $g$ is entire and bounded away from the real line, so must be constant.
A: 
$$f(z)= \dfrac{\sin(z)}{\sin(\pi z)} -\frac1\pi \lim_{N\to \infty}\sum_{0<|n|\le N}(-1)^n \dfrac{\sin(n)}{z-n}=g(z)-h(z)$$ is entire.

Differentiating, $g'$ is bounded away from its poles, the series for $h'$
is uniformly convergent thus bounded away from its poles, and the boundedness of $f'$ on $|z-n|=1/2$ plus the Cauchy integral formula implies that $f'$ is bounded on $|z-n|<1/3$, thus constant.
Since $g'(i\infty)=h'(i\infty)=0$ it means that $f' = 0$.
Thus $f$ is constant ie. $$f = f(0) =\frac1\pi $$
The general Mittag-Leffler theorem is mostly the abstraction of this reasonning.
