Laurent Series of $f(z) = \frac{1}{e^z - 1}$

Question

The Laurent Series of $f(z)$ centred at $0$ can be written as,

$$f(z) = \frac{1}{z} - \frac{1}{2} + \frac{z}{12} - \frac{z^3}{720} + \cdots$$

So we see that $f(z)$ has a simple pole at $0$. Can we conclude that the ring of convergence of the Laurent Series is the exterior of the unit disk $|z|\geq 1$ and if so, what is the radius of convergence?

Answer 1

No, it converges on $\{0<|z|<2\pi\}$, since the next pole occurs at $e^{\pm 2\pi i}$.