Maclaurin series of $x^3/(e^x-1)$ how would i taylor expand $f(x)=\frac{x^3}{e^x-1}$ around $x=0$?
I was thinking of writing
$\frac{x^3}{e^x-1}\approx\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots}$
$~~~~~~~~= \frac{1}{\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{2x}+\frac{1}{6}+\frac{x}{24}+\frac{x^2}{120}+\dots}$
$~~~~~~~~=6\bigg(\frac{1}{1+(\frac{6}{x^3}+\frac{6}{x^2}+\frac{3}{x}+\frac{x}{4}+\frac{x^2}{20}+\dots)}\bigg)$
And then use the geometric series $\frac{1}{1+x}=1-x+x^2-x^3+\dots$ but that didn't get ge the right answer..

Do you probably know that $$\frac{x}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^n}{n!},$$ where $B_n$ are the Bernoulli's numbers. Hence $$\frac{x^3}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^{n+2}}{n!}=x^2-\frac{x^3}{2}+\frac{x^4}{12}+ \dots.$$
