Taylor expansion and coefficients of $f(z) = \frac{z}{e^z-1}$. This is a problem from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming qual. I thought this one would be fairly straightforward, but since it's a qual question... of course it isn't. Below is the question and what I've tried.
Problem:
Consider the function
$$
f(z) = \frac{z}{e^z-1}.
$$
(a) Find the first four terms of the Taylor expansion about $z=0$ and the radius of convergence. 
(b) Find all coefficients expressed through recurrent relation (this can count as first part of a solution of part (a)).
What I've considered:
First, I believe that a Taylor series about zero for this function would have a radius of convergence up to the first non-analytic point (aside from zero). This would occur where $e^z = 1$, thus at $z=2\pi i$.
Now, I've tried starting with the Taylor series definition
$$
f(a) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(z-a)^k,
$$
For $f(0)$ have that $f(0) = \frac{0}{0}$, so I instead take the limit
$$
f(0) = \lim_{z\to 0} \frac{z}{e^z-1} = \lim_{z\to 0} \frac{1}{e^z} = 1.
$$
My first coefficient in a Taylor series $f(a) = \sum_{k=0}^\infty a_kz^k$ is thus $a_0 = 1$. Beyond this, I run into trouble. The derivatives of $f(z)$ all seem to evaluate to infinity when $z=0$. So I tried something else.
Consider $f(z) = -z\cdot \frac{1}{1-Q(z)}$, where $Q(z) = 1 + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \dots $. Then using the geometric series, for $|Q(z)|<1$, we have
$$
f(z) = -z \sum_{k=0}^\infty Q(z)^k.
$$
The thought here was to work out the first four (or more if needed?) terms algebraically and then see if I can find a relationship such that I can specify all coefficients. However, the first term of $1$ in $Q(z)$ seems to be problematic, in that I seem to end up without a finite value for each term... every expanded term has a $z$ term, for example.
Finally, I considered the fact that $e^z-1$ seems to remove said problematic $1$. One can also cancel the $z$ in the numerator, such that we'd have
$$
f(z) = \frac{z}{z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \dots} = \frac{1}{1 + \frac{z}{2!} + \frac{z^2}{3!} + \frac{z^3}{4!} + \dots},
$$
but I can't see anywhere useful to go from here. If someone can please walk me through how to do this problem, I would very much appreciate it. I've been fighting with it for a while.
 A: You can find this question in page 97 of stein's fourier book (the 4th problem of the third chapter ) , I give you that question and my solution as follows:
4.In this problem, we find the formula for the sum of the series
$$\sum\limits_{0}^{\infty}\dfrac{1}{n^k}$$
where k is any even integer.These sums are expressed in terms of the Bernoulli numbers;the related Bernoulli polynomials are discussed in the next problem.
Define the Bernoulli numbers $B_n$ by the formula
$$ \dfrac{z}{e^z-1}=\sum\limits_{n=0}^{\infty}\dfrac{B_n}{n!}z^n, $$
$(a)$ Show that $B_0=1,B_1=-\dfrac{1}{2},B_2=\dfrac{1}{6},B_3=0,B_4=-\dfrac{1}{30}$,and $ B_5=0$.
$(b)$ Show that for $n \geq 1$ we have
$$ B_n=\dfrac{1}{n+1}\sum\limits_{k=0}^{n-1}\binom{n+1}{k}B_k. $$
My solution is as follows:
$(a)$ $1=\dfrac{z}{e^z-1}\dfrac{e^z-1}{z}=(\sum\limits_{n=0}^{\infty}\dfrac{B_n}{n!}z^n)(\sum\limits_{n=0}^{\infty}\dfrac{z^n}{(n+1)!})=\sum\limits_{n=0}^{\infty}(\sum\limits_{k=0}^{n}\dfrac{B_k}{k!(n+1-k)!})z^n$
we can get $\sum\limits_{k=0}^{n}\dfrac{B_k}{k!(n+1-k)!}=0$,and it's easy to see $B_0=1$,we could use the formula on the left to get our conclusion.
$(b)$ We know $\sum\limits_{k=0}^{n}\dfrac{B_k}{k!(n+1-k)!}=0$ from (a),and we could get $ B_n=\dfrac{1}{n+1}\sum\limits_{k=0}^{n-1}\binom{n+1}{k}B_k. $
A: Part a and b are both immediately solved using Dividing an infinite power series by another infinite power series
The above method shows that in general, if you know the power series for $f(z)$ and $g(z)$ and you know that $g(0) \neq 0$, then the power series coefficients of $h(z) = f(z)/g(z)$ are immediately obtained as a recursive formula in terms of the coefficients of $f$ and $g$.
A: The coefficients are called the Bernoulli numbers. My guess is that people who wrote the question expect that you are familiar with them.
