What is the radius of convergence of $\frac{\sinh(z)}z$ about $z=\pi$

Question

I have a question that asks to find the radius of convergence for $\frac{sinh(z)}z$ about $z=\pi$. I attempted to solve this as: $$\frac{d}{dx} \frac{\sinh(z)}z = \frac{\cosh(z)}z - \frac{\sinh(z)}{z^2}$$ which isn't defined at $z=0$, therefore the radius of convergence would be: $$R=\lvert{0-pi}\lvert = \pi$$ is this correct?

Answer 1

Hint: $$\sinh(z)=\frac12\left(e^z-e^{-z}\right)=\sum_{k=0}^\infty\frac{z^{2k+1}}{(2k+1)!}$$

So what what is the nature of $\frac{\sinh z}{z}$ at $z=0?$