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