# Maclaurin Series of $\frac{1}{e^x -1}$

I want to find the MacLaurin Series for the function $f(x) = \frac{1}{e^x -1}$. But when I compute the first derivative of $f(x)$: $$\frac{d}{dx}\frac{1}{e^x -1} = -\frac{e^x}{(e^x-1)^2}$$

A the point $x=0$, I get an indeterminate expansion: $$f'(0)=-\frac{e^0}{(e^0-1)^2}$$

So how can I compute the series for this function $f(x)$?

• Consider a geometric series.
– user61527
Commented Feb 19, 2014 at 6:22
• A related problem. Consider the function $x/(e^{x}-1)$. Commented Feb 19, 2014 at 6:39
• @T. Bongers : That is probably a very bad idea, considering the geometric series doesn't even converge at $0$. Commented Feb 19, 2014 at 6:43
• @PatrickDaSilva In a sense it is a very good idea, as it will give us the best thing we can hope for - the Laurent expansion. It's agreed though the OP needs to be informed there is no Maclaurin expansion.
– anon
Commented Feb 19, 2014 at 6:55
• @PatrickDaSilva: A major part of learning mathematics is to allow oneself to be exposed to ideas that generalize and extend concepts we are already familiar with. To that end, simply saying "the Maclaurin series is not defined" and to stop there, is not conducive to learning. If you are satisfied with that answer, so be it: but your opinion should not preclude others from giving a more detailed response. After all, it is not as if we are pretending that a Maclaurin series exists. We clearly stated otherwise and explained why. Commented Feb 19, 2014 at 7:09

Your Series hasn't a MacLaurin expansion at $x=0$, since its undefined at this point, but you can find a Laurent expansion for it as follows. Note that $$\frac{e^x-1}x=\sum_{n=0}^\infty\frac1{(n+1)!}x^n$$ By Power Series Division Theorem, the quotient $\frac1{\frac{e^x-1}x}=\frac x{e^x-1}$ also has a power series expansion near $x=0$. It is customary to denote its coefficients by $\frac{B_n}{n!}$, in which case we can write $$\frac x{e^x-1}=\sum_{n=0}^\infty\frac{B_n}{n!}x^n$$ Hence, $$\frac 1{e^x-1}=\sum_{n=0}^\infty\frac{B_n}{n!}x^{n-1},x\neq0$$ The numbers $B_n$ are called the Bernoulli numbers. Also see here to calculate these numbers.

$f$ is not even defined at $x = 0$. Notice that $e^0 = 1$, hence $\frac 1{e^0 - 1}$ is not defined, at least not by the expression. Note that $$\lim_{x \to 0^+} e^x - 1 = 0^+ \Rightarrow \lim_{x \to 0^+} \frac 1{e^x - 1} = +\infty, \quad \lim_{x \to 0^-} e^x - 1 = 0^- \Rightarrow \lim_{x \to 0^-} \frac 1{e^x - 1} = -\infty,$$ so $f$ is not continuous at $0$ in any possible way. Its derivative doesn't exist, thus a MacLaurin expansion is out of the question.

That pretty much explains why you couldn't compute it : it's because -- you can't --.

Hope that helps,

• It's worth mentioning that $\frac{x}{e^x-1}$ continuously extends to $x=0$, and that extension has a Maclaurin series famously involving Bernoulli numbers. Considering that, $\frac{1}{e^x-1}$ does have a Laurent series centered at $0$. Commented Feb 19, 2014 at 7:04
• I also think that it's reasonable to refer to a Laurent seires centered at $0$ as a Maclaurin series. A Maclaurin series is a Taylor series centered at $0$. It's the "centered at $0$" part that gives it the special name Maclaurin. So while it would be a little nonstandard, it's not a crazy thing to call this function's Laurent series a Maclaurin series. Commented Feb 19, 2014 at 7:08
• @alex.jordan : Laurent series are called Laurent series to distinguish them from MacLaurin series. So if you want to say that you compute the MacLaurin series exists, it should not be a Laurent series of a function which admits no MacLaurin series, otherwise we invented terms for no reason. But discussing them never hurts, of course! Commented Jan 9, 2016 at 1:41
• For me, the distinguishing feature of a Laurent series (distinguishing it from a power series or a Taylor series) is that it allows for some negative powers. And that is what justified the new name for a new object. So I've never seen it's main purpose as being something that is distinguished from Maclaurin series, even though that is an automatic consequence of the way I see it too. Anyway, just commenting on where I was coming from. Commented Jan 9, 2016 at 2:48
• I think when answering Thiago’s question we need to think of the likelihood that either (a) they’re a beginner and confused MacLaurin and Laurent series or (b) they know exactly what a MacLaurin series is, but for some reason didn’t know the basic property that it doesn’t exist when f(0) isn’t defined. As someone coming from a physics background (a) is a lot more likely Commented Oct 30, 2022 at 2:53

Technically, the series expansion about $x = 0$ of $f(x) = (e^x - 1)^{-1}$ is not a Maclaurin series, because the function is not defined at $x = 0$. Therefore, a series expansion of this function must have a term of the form $1/x$, and is a Laurent series.

To find the series expansion, consider the following definition: Let $\{B_n\}_{n \ge 0}$ be a sequence of numbers such that $$\sum_{k=0}^n \binom{n}{k} B_k = \begin{cases} B_n & n \ne 1, \\ B_1 + 1, & n = 1. \end{cases}$$ This sum is the binomial convolution of the sequences $\{B_n\}$ and $\{1\}$; i.e., if $h_n = \sum_{k=0}^n \binom{n}{k} B_k$, then $$\sum_{n=0}^\infty h_n \frac{z^n}{n!} = \sum_{k=0}^\infty B_k \frac{z^k}{k!} \sum_{j=0}^\infty \frac{z^j}{j!} = e^z \sum_{k=0}^\infty B_k \frac{z^k}{k!} = e^z \hat B(z),$$ where $\hat B(z) = \sum_{k=0}^\infty B_k \frac{z^k}{k!}$. But the right-hand side has exponential generating function $$B_0 \frac{z^0}{0!} + (B_1 + 1) \frac{z^1}{1!} + \sum_{n=2}^\infty B_n \frac{z^n}{n!} = z + \sum_{n=0}^\infty B_n \frac{z^n}{n!} = z + \hat B(z).$$ Therefore, $z + \hat B(z) = e^z \hat B(z)$, and $$\hat B(z) = \frac{z}{e^z-1}.$$ Dividing both sides by $z$ gives the desired series expansion. Explicitly, we have $$f(x) = \frac{1}{x}-\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}-\frac{x^7}{1209600}+\frac{x^9}{47900160}+O(x^{11}).$$

• Is there a closed form for the coefficients in the Laurent series? Commented Nov 4, 2019 at 4:35
• fun fact this series proves that $1 + 1 + 1 ... = -\frac{1}{2}$ and $1 + 2 + 3 + ... = -\frac{1}{12}$ and the rest of $\zeta(-2n+1) = \text{bernoulli}$ Commented Oct 4, 2022 at 19:36