# Find the series expansion of $\frac{e^z - 1}{z}$ about zero and find its radius of convergence.

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?

• There is an important characterization of the radius of convergence of power series expansion around $x_0$ in terms of the distance to the nearest singularity to $x_0$. Is that a topic you've covered? Nov 23, 2018 at 15:03
• It does not come to mind immediately, but I am returning to this material from a previous semester and may have just forgotten the topic.
– Zed1
Nov 23, 2018 at 15:06
• Another approach would be to use the ratio test on the power series presented in the Answers to your linked Question. Nov 23, 2018 at 15:09

The radius of convergence is either a non-negative real number or $$\infty$$. In this case, it turns out to be $$\infty$$, because$$\frac{e^z-1}z=1+\frac z2+\frac{z^2}{3!}+\frac{z^3}{4!}+\cdots$$and this series converges (absolutely) everywhere, by the ratio test.
• Just to clarify, manipulating the series $e^z = \sum_{k=0}^{\infty}{\frac{z^k}{k!}}$ by subtracting the series by 1 and dividing each term by $z$ to obtain $\frac{e^z - 1}{z}$ is perfectly valid?