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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?

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  • $\begingroup$ 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? $\endgroup$
    – hardmath
    Nov 23, 2018 at 15:03
  • $\begingroup$ 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. $\endgroup$
    – Zed1
    Nov 23, 2018 at 15:06
  • $\begingroup$ Another approach would be to use the ratio test on the power series presented in the Answers to your linked Question. $\endgroup$
    – hardmath
    Nov 23, 2018 at 15:09

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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.

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    $\begingroup$ 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? $\endgroup$
    – Zed1
    Nov 23, 2018 at 15:17
  • $\begingroup$ Perfectly. You can do it whenever you want. $\endgroup$
    – José Carlos Santos
    Nov 23, 2018 at 16:52
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    $\begingroup$ @Zed1 You should validate old answers when they are satisfying like this one. I have had a look at all the answers that were given to you : you have validated rather few of them. Think that it is a way to maintain this site in good order... and as well satisfying for people who have spent sometimes hours to answer in a convenient way. $\endgroup$
    – Jean Marie
    Mar 15, 2020 at 18:53

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