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 $f(x)$?
 A: 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.
A: $f$ 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,
A: 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 $$f(x) = \frac{1}{x}-\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}-\frac{x^7}{1209600}+\frac{x^9}{47900160}+O(x^{11}).$$
