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.
 A: Let be
$$\sum_{n=0}^{\infty}a_nz^n=\frac{z}{e^z-1}.$$
Then,
$$z=\left(\sum_{n=1}^\infty \frac{z^n}{n!}\right)
\left(\sum_{n=0}^{\infty}a_nz^n\right)=a_0z+\cdots$$
A: Hint
Why don't apply the definition of the series expansion $$f(z)=f(0)+\frac{z}{1!}f'(0)+\frac{z^2}{2!}f''(0)+\frac{z^3}{3!}f'''(0)+\frac{z^4}{4!}f''''(0)+\cdots$$ Using $$f(z)=\frac{z}{e^z-1}$$ you should arrive quickly (using limits) to $f(0)=1$, $f'(0)=-\frac{1}{2}$, $f''(0)=\frac{1}{6}$, $f'''(0)=0$,  $f''''(0)=-\frac{1}{30}$.
Another way is lond division and another is what Martín-Blas Pérez Pinilla suggests (probably the fastest).
A: $$\frac{z}{e^z - 1}=\frac{1}{1+S} = 1 - S + S^2 - S^3 + \ldots$$
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 
$$1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^2 -\\- (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^3  +(\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} )^4= \\=
1 - (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24}+ \frac{z^4}{120} ) + (\frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24})^2 - (\frac{z}{2} + \frac{z^2}{6}  )^3 + (\frac{z}{2} )^4 + \text{h.o.t.} = \\ = 1 - \frac{z}{2} +\frac{z^2}{12} -\frac{z^4}{720} + \text{h.o.t.}$$
A: For exponential powerseries $A(z)=\displaystyle\sum_{n\geqslant 0} a_n \dfrac{z^n}{n!}$, $B(z)=\displaystyle\sum_{n\geqslant 0} b_n\dfrac{z^n}{n!}$, under appropriate convergence assumptions, we have that $C(z)=A(z)B(z)=\displaystyle\sum_{n\geqslant 0} c_n\dfrac{z^n}{n!}$ where $\displaystyle c_n=\sum_{k=0}^n\binom nka_nb_{n-k}$. In your case, for $A(z)=\dfrac{z}{e^z-1}$ and $B(z)=\dfrac{e^z-1}{z}$, we have that $a_n=\dfrac{1}{n+1}$, $c_n=[n=0]$, so that the coefficients $b_n$ are determined by $b_0=1$ and by the recursive formula $$\sum_{k=0}^n \binom nk \frac{1}{k+1}b_{n-k}=0\;,n>0$$
These are the Bernoulli numbers.
