<s>[INST] First four terms of the power series of $f(z) = \frac{z}{e^z-1}$? Attempt:
$$ e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$
$$ e^z - 1 = \sum_{n=0}^\infty \frac{z^n}{n!} -1$$
$$ e^z - 1 = z\sum_{n=0}^\infty \frac{z^n}{(n+1)!} $$
Thus
$$  \frac{z}{e^z-1} = \frac{1}{\sum_{n=0}^\infty \frac{z^n}{(n+1)!}}.$$
How to proceed from this form? $z$ is complex.
 [/INST] $$\frac{z}{e^z - 1}=\frac{1}{1+S} = 1 - S + S^2 - S^3 + \ldots$$
where $S = \sum_{n=1}^{\infty} \frac{z^n}{(n+1)!}$
To get the first three terms in the series expansion at $z=0$  consider the first three terms in the series expansion of 
$1 - \bar S + \bar S^2 $ where $\bar S =  \sum_{n=1}^{2} \frac{z^n}{(n+1)!} = \frac{z}{2} + \frac{z^2}{6} $, that is 
$$1 - (\frac{z}{2} + \frac{z^2}{6}) + (\frac{z}{2} + \frac{z^2}{6})^2 = 1 -\frac{z}{2} + \frac{z^2}{12} + \text{h.o.t.}$$ 
To get the first $5$ terms consider 
$$1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^2 -\\- (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^3  +(\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^4= \\=
1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24})^2 - (\frac{z}{2} + \frac{z^2}{6}  )^3 + (\frac{z}{2} )^4 + \text{h.o.t.} = \\ = 1 - \frac{z}{2} +\frac{z^2}{12} -\frac{z^4}{720} + \text{h.o.t.}$$
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