Solution
= Solution
The removable value at zero is one. Expand the denominator as $e^x-1=x(1+x/2+x^2/6+\cdots)$ and determine the reciprocal coefficients by multiplication. The first four nonzero terms are
$$
\boxed{\frac{x}{e^x-1}=1-\frac x2+\frac{x^2}{12}-\frac{x^4}{720}+O(x^6).}
$$
This is the <exponential generating function> of the <Bernoulli numbers>, with convention $B_1=-1/2$. The coefficients of $x^3$ and $x^5$ are zero.