Series for $\sin(z) / \sin(\pi z)$ ${\sin(z) \over \sin(\pi z)} = 1/\pi + {z \over \pi} \sum_{n \in {\bf Z} \setminus \{0\}} {(-1)^n \sin(n) \over n(z-n)}$

First I apply  Mittag-Leffler's theorem, to see that the RHS is a mereomorphic function with the appropriate residues and orders for each pole, thus the LHS and RHS differ by an analytic function.

How can I then show that this analytic function is identically 0?  One idea is to use Liouville's theorem to show it it bounded; I do not see how to do that.

We do not have periodicity on either side.
This is equivalent to showing that $$\frac{sin(z)}{sin(\pi z)}=\frac{2}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{ n sin(n)}{z^2-n^2}$$
Again, Mittag-Lefler says these functions differ by a holomorphic function. You can take the derivative on both sides, and then the left and right side become periodic. At this point, you can prove that the derivative of the holomorphic function they differ by is zero, therefore, the function itself is constant. Finally, look at the value at $z=0$ on both sides (noting that the left side has a removable singularity at $0$). This tells you that the constant must be $0$.
This is equivalent to showing that $$\frac{sin(z)}{sin(\pi z)}=\frac{2}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{ n sin(n)}{z^2-n^2}$$
Again, Mittag-Lefler says these functions differ by a holomorphic function. You can take the derivative on both sides, and then the left and right side become periodic. At this point, you can prove that the derivative of the holomorphic function they differ by is zero, therefore, the function itself is constant. Finally, look at the value at $z=0$ on both sides (noting that the left side has a removable singularity at $0$). This tells you that the constant must be $0$.