Series expansion for $f(x)=\frac{x}{e^x-1}$ I am trying to expand the function $f(x)=\dfrac{x}{e^x-1}$ into Maclaurin series and I encounter this problem.
The first derivative is $f'(x)=-\dfrac{e^x(x-1)+1}{(e^x-1)^2}$. When $f(0)=\dfrac{1(0-1)+1}{(1-1)}^2$, which is undefined. Can I continue to expand the series using the second derivative.
Is there a faster way to expand this series without directly employing the Maclaurin formula?
According to this page:
https://proofwiki.org/wiki/Definition:Bernoulli_Numbers
The series is :
$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}+\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$
 where $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0$
How can I expand the series as is shown in that page?
 A: $\dfrac{e^x-1}x=\dfrac{x+\dfrac{x^2}2+\dfrac{x^3}6+...}x=1+\dfrac12x+\dfrac16x^2...$
$\therefore\left(B_0+B_1x+B_2\dfrac{x^2}2+...\right)
\left(1+\dfrac12x+\dfrac16x^2...\right)=1$
$\therefore B_0+\left(B_1+\dfrac12B_0\right)x+\left(\dfrac{1}2B_2+\dfrac12B_1+\dfrac16B_0\right)x^2+...=1$
$\therefore B_0=1, B_1+\dfrac12B_0=0, \dfrac12B_2+\dfrac12B_1+\dfrac16B_0=0, ...$
$\therefore B_0=1, B_1=-\dfrac12, B_2=\dfrac16, ...$
A: One (not aware of the Bernoulli numbers) can compute the reciprocal series for $$\frac{e^x-1}{x}=\sum_{n=0}^{\infty}a_n x^n,\qquad a_n=\frac{1}{(n+1)!}.$$ That is, assuming $x/(e^x-1)=\sum_{n=0}^{\infty}b_n x^n$, we get the system of equations for unknowns $b_n$ from $\left(\sum_{n=0}^{\infty}a_n x^n\right)\left(\sum_{n=0}^{\infty}b_n x^n\right)=1$, i.e. \begin{align}\color{gray}{[x^0]}&\quad a_0 b_0=1,\\\color{gray}{[x^1]}&\quad a_0 b_1+a_1 b_0=0,\\\color{gray}{[x^2]}&\quad a_0 b_2+a_1 b_1+a_2 b_0=0,\\\color{gray}{[x^3]}&\quad a_0 b_3+a_1 b_2+a_2 b_1+a_3 b_0=0,\\&\ldots\end{align} which gives \begin{align}b_0&=1/a_0=1,\\b_1&=-a_1 b_0/a_0=-1/2,\\b_2&=-(a_2 b_0+a_1 b_1)/a_0=1/12,\\b_3&=-(a_3 b_0+a_2 b_1+a_1 b_2)/a_0=0,\\b_4&=-(a_4 b_0+a_3 b_1+a_2 b_2+a_1 b_3)/a_0=-1/720,\end{align} and so on.
