<s>[INST] Maclaurin series for $\frac{x}{e^x-1}$ Maclaurin series for 
$$\frac{x}{e^x-1}$$
The answer is 
$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$
How can i get that answer?
 [/INST] One way is to write $e^x-1 $ as $1 + x + x^2/2 + ... - 1$ and then factor out $x$ and cancel up the top and expand it as geometric series and collect the coefficients of like powers. 
$\displaystyle 
\begin{align*}
e^x - 1 &= x + \frac{x^2}{2!} + \frac{x^3}{3!} + o(x^4)\\ 
\frac{x}{e^x - 1} &=  \frac{1}{1 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )} \\
&= 1 - \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right ) + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^2 - \left( \frac x 2 - \frac{x^2}{6} + o(x^3) \right )^3 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^4... \\
&= 1 - \frac{x}{2} + x^2 \left( \frac 1 4 - \frac 1 6  \right ) + x^3 \left(-\frac{1}{4!} + 2 \cdot  \frac 1 2 \cdot \frac 1 6 - \frac{1}{2^3} \right ) + x^4 \left(-\frac{1}{5!} + \frac{1}{6^2}  + 2 \cdot \frac 12 \cdot \frac{1}{4!} + \frac{1}{2^4}\right )+o(x^5)
\end{align*} $
</s>