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How to expand the series $\dfrac{1}{e^x-1}$

Series expansion for $f(x)=\frac{x}{e^x-1}$

In these topics, I have asked for a way to develop the function $\dfrac{x}{e^x-1}$ into series. I won't do that here. From what I see, this function should not have a Maclaurin series (centered at 0) since the function and its higher derivative remain undefined when x=0. Then what is the name for the series derived from the topics that I mention above. Is it a Laurent series or Puisseux series?

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    $\begingroup$ Technically $f(x) = x/ (e^x - 1)$ is not defined at $x=0$, but the limit exist as $x \rightarrow 0$, so it is a removable discontinuity. If you define a new function $g(x) = x/(e^x-1)$ for $x \neq 0$, and $g(0) = 1$, then the function has a Maclaurin series. $\endgroup$ – Jair Taylor Nov 6 '19 at 18:41
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The function itself is not a series, of any kind. A series is one way to represent a function.

I think the term you're looking for is Laurent series. This is the generalisation of Maclaurin series that allows for negative integer exponents.

Puisseux series allow for fractional exponents as well. That is not what these examples use, so while the term may technically be applicable (I don't know much about them personally), it's not what I would call it.

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  • $\begingroup$ But can I form a Laurent series to represent a function when it is not defined at 0? $\endgroup$ – James Warthington Nov 6 '19 at 18:21
  • $\begingroup$ Yes. That's the point. Laurent series, in addition to terms like $x^2$ and $x^{739}$ also has terms like $\frac1x$ and $\frac1{x^{1337}}$. Those are not defined at $0$, but they are defined around $0$. $\endgroup$ – Arthur Nov 6 '19 at 18:47

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