Characterize all singularities of $$z\mapsto\frac{z}{\mathrm e^z-1}.$$
What is the radius of convergence of the taylor expansion of $$\frac{z}{\mathrm e^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}$$ and deduce whether the $B_n$ are bounded?
Obviously $z=2n\pi\mathrm i$ for all $n\in\mathbb Z$ are singularities. First let's take a look at $z=2n\pi\mathrm i$ for all $n\in\mathbb Z\setminus\{0\}$. From l'Hôpital we obtain
$$ \lim_{z\to 2n\pi\mathrm i}\frac{(z-2n\pi\mathrm i)z}{\mathrm e^z-1} = \lim_{z\to 2n\pi\mathrm i}\frac{2z - 2n\pi\mathrm i}{\mathrm e^z} = \lim_{z\to 2n\pi\mathrm i}\frac{2}{\mathrm e^z} = 2 $$
which is finite hence for all $n\in\mathbb Z\setminus\{0\}$ those $z$ are simple poles.
But what about $z=0$? Can this same argument be applied to this as well?
(Edit. What about analytic continuation - is it possible for $z=0$?)
Furthermore from Cauchy's differentiation formula it follows with $\gamma=\partial B_\rho(0)$ that $$B_n=\frac{n!}{2\pi\mathrm i}\oint_{\gamma}\frac{z}{\mathrm e^z-1}\cdot\frac{\mathrm dz}{z^{n+1}}$$ with $\rho\in(0,2\pi)$ since the curve would otherwise enclose one or more singularities. Therefore I assume that the radius is at most $2\pi$ and therefore all $B_n$ bounded as well. Unfortunately I did not come up with a proper proof why this may be true.
Since $$ \lim_{z\to0}\frac{e^z-1}{z}=1 $$ you also have $$ \lim_{z\to0}\frac{z}{e^z-1}=1 $$ so the point $0$ is a removable singularity for $f$; call $\hat{f}$ the extended function. Since $2\pi i$ is the nearest singularity to $0$ for $\hat{f}$ and it is a pole, we know that the Taylor series at $0$ has radius of convergence $|2\pi i|=2\pi$.
The points $2n\pi i$ ($n\ne0$) are simple poles for $f$, but you're computing the residue wrongly: $$ \lim_{z\to 2n\pi i}\frac{z(z-2n\pi i)}{e^z-1}= \lim_{z\to 2n\pi i}\frac{2(z-n\pi i)}{e^z}=2n\pi i $$ (you can't further apply l'Hôpital). Alternatively, use the substitution $z=w+2n\pi i$, so the limit is $$ \lim_{z\to 2n\pi i}\frac{z(z-2n\pi i)}{e^z-1}= \lim_{w\to 0}\frac{w}{e^w-1}(w+2n\pi i) $$