# What kind of series is the function $\frac{x}{e^x-1}$?

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?

• 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. – Jair Taylor Nov 6 '19 at 18:41

• 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 $0$. – Arthur Nov 6 '19 at 18:47