Find the series expansion of $$\frac{e^z - 1}{z}$$ about zero and find its radius of convergence.
Part of this question was asked here, Finding the Taylor series expansion of $f(z)=\frac{e^{z}-1}{z}$ around $0$, but they do not mention anything about the radius of convergence.
Since the function is analytic everywhere except at the singularity at z = 0, would the radius of convergence be $$ 0 < \left|z \right| < \infty? $$
I'm also not fully convinced that I am just able to do the obvious thing of subtracting the series expansion of $e^z$ by 1 and then dividing by $z$.
Does dividing by $z$ not cause any problems, besides restricting the radius of convergence?