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..
 A: If you assume it has a Taylor expansion, you can write:
$$\frac{x^3}{e^x-1} = a_0+a_1x+a_2x^2+a_3x^3+\ldots$$
Rearrange:
$$x^3 = \left( a_0+a_1x+a_2x^2+a_3x^3+\ldots \right)(e^x-1) $$
Use the series of $e^x$:
$$x^3 = \left( a_0+a_1x+a_2x^2+a_3x^3+\ldots \right)\left( x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24} + \ldots \right) $$
Expand and group in powers of $x$:
$$x^3 = a_0x+ \left( \frac{a_0}{2}+a_1 \right)x^2 + \left( \frac{a_0}{6}+  \frac{a_1}{2}+a_2 \right)x^3 + \left( \frac{a_0}{24}+  \frac{a_1}{6}+\frac{a_2}{2} + a_3 \right)x^4 + \ldots$$
Identifying the coefficients of the corresponding powers of $x$ then allows you to find the coefficients $a_0,a_1,a_2,a_3,\ldots$ recursively.
$$\left\{
\begin{array}{l}
a_0 = 0 \\
\frac{a_0}{2}+a_1 = 0 \\
\frac{a_0}{6}+  \frac{a_1}{2}+a_2 = 1 \\
\frac{a_0}{24}+  \frac{a_1}{6}+\frac{a_2}{2} + a_3 = 0 \\
\cdots
\end{array}
\right. \Rightarrow \left\{
\begin{array}{l}
a_0 = 0 \\
a_1 = 0 \\
a_2 = 1 \\
a_3 = -\frac{1}{2} \\
\cdots
\end{array}
\right.$$
Tedious and perhaps not that elegant, but it works!
$$\frac{x^3}{e^x-1} = x^2-\frac{x^3}{2}+\frac{x^4}{12}+\ldots$$
A: 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 $$\frac{x^3}{e^x-1} = \sum_{n=0}^{+\infty} B_n \frac{x^{n+2}}{n!}=x^2-\frac{x^3}{2}+\frac{x^4}{12}+ \dots.$$
