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I'm trying to find the Power Series expansion and Radius of Convergence for the function $$f(z) = \frac{z}{e^{z} - 1}$$ at points $z = 0$ and $z = 1$.

When it comes to the radii of convergence, $e^{i\cdot(2n\pi)} = 1$ so the function has poles on the imaginary axis at intervals of $2\pi$. So, the series expansion centered at zero has radius of convergence equal to $2\pi$, whereas the series centered at $z=1$ has radius $1$ since the closest pole lies at $z = 0$.

Now, when it comes to actually finding the series expansion, I'm unsure how to find the $n^{th}$ derivative. How does one do so? And, is there a cleverer way of finding the series coefficients without computing the $n^{th}$ derivative?

kdeoskar
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  • hint: geometric series – Sine of the Time Dec 26 '23 at 15:17
  • @SineoftheTime Thank you for the hint! I tried proceeding by expanding $e^{z}$ and simplifying to $\frac{z}{e^z - 1} = \frac{1}{1+P(z)} = P(z) + P(z)^2 + P(z)^3 + \cdots$ where $P(z) = \frac{z}{2!} + \frac{z^2}{3!} + \dots$ (obtained from the earlier simplification)

    But this only holds for $|P(z)| < 1$. I'm not sure how to show that this is true. Do you have any suggestions for doing this without integration? (My textbook gives this question before teaching integration).

    – kdeoskar Dec 26 '23 at 15:46
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    look up Bernoulli numbers as $\frac{z}{e^{z} - 1}$ is their generating function – Conrad Dec 26 '23 at 15:51
  • @KeshavBalwantDeoskar how would you do it using integration? – Sine of the Time Dec 26 '23 at 16:07
  • @SineoftheTime An answer to this question asking for the Laurent Series of $1/e^{z} - 1$ used integration to find the coefficients so I figured I'd mention I haven't studied that yet in case it's a standard method – kdeoskar Dec 26 '23 at 17:20
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    oh, you mean by the def of the coefficients. This is usually not done, you use expansion of well known series. Related – Sine of the Time Dec 26 '23 at 18:07
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    @SineoftheTime I see! I got it. Thanks again for the help – kdeoskar Dec 28 '23 at 13:45

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