Show that a given function is entire

Question

I've stumbled across this problem. Q: Show that $f(z)=\sum_{n=-\infty}^{\infty} \frac{\sin z}{|n|! (z - n\pi)}$ is an entire function, and evaluate $\sum_{n=0}^{\infty} f(n \pi).$ Any suggestions is much appreciated.

Answer 1

Hint:

Fix an integer $n$. Consider the power series expansion of $\sin(z)$ about $n\pi$: $$\sin(z)=\sum_{k=0}^\infty (-1)^k \frac{(z-n\pi)^{2k+1}}{(2k+1)!}=(z-n\pi)\sum_{k=0}^\infty(-1)^k\frac{(z-n\pi)^{2k}}{(2k+1)!}=(z-n\pi)\varphi_n(z)$$

for an entire $\varphi_n$.

This shows that the partial sums of your series are entire. Recall also that if a sequence of holomorphic (say on $U$) functions converges uniformly on compact sets, then the limit function is also holomorphic on $U$.