<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] An other approach would be to take the inverted first:
$$ \frac{e^x-1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} + \frac{x^4}{120} \cdots $$
and then you go:
$$\frac{x}{e^x-1} = (\frac{e^x-1}{x})^{-1} = [1 + (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120})]^{-1} $$
and then expand this as in the binomial expansion $(1+u)^{-1}$ and since here $ u <1$ the previous gets expanded to:
$$ 1 - (\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24} +\frac{x^4}{120}) +(\frac{x}{2} + \frac{x^2}{6} +\frac{x^3}{24})^2 - (\frac{x}{2} + \frac{x^2}{6})^3 + (\frac{x}{2})^4 \tag1 $$ Notice that i only take terms accordingly until the desired $x^4$ terms neglecting the other higher order terms. Now keep expanding $(1)$ while still not taking into account calculations that will result in terms higher than $x^4$ (to avoid unnecessary trouble) and it won't be long until you reach the correct result.
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