nth root of Bernoulli numbers I'm trying to prove that the supremum limit is equal to infinity:
$\limsup_{n->\infty}\sqrt[n]{|B_n|}=\infty$
Where $B_n$ is defined via the series expansion:
$f(z)=\frac{z}{e^{z}-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}z^n$
Initially, I thought that I could use the nth derivative of f(z) at zero since:
$f^{(n)}(0)=\frac{B_n}{n!}$ 
But I can't find a general formula for the nth derivative. I also tried to use the series expansion for $f(z)*(e^z-1)=\sum_{n=1}^\infty z^nn!\sum_{k=0}^{n-1}\binom{n}{k}B_k$ but I'm confused about this since the left hand side equals to z and so the nth derivative (n>2) is just equal to zero.
I've run out of ideas, so any tips would be greatly appreciated!
 A: The radius of convergence of the power series of $f$, centered at $0$, is $2\pi$ ($2i\pi$ is the closest point at the origin, where $f$ has a singular point). So, the limsup formula for the radius of convergence yields that
$$\limsup_{n\rightarrow +\infty}{\left(\frac{|B_n|}{n!}\right)^{\frac{1}{n}}}=\frac{1}{2\pi}.$$
Consequently, for $\epsilon=1/4\pi,$ there exists $n_0\in \mathbb{N}^*$ such that $\forall n\geq n_0$ it is true that 
$$\left(\frac{|B_n|}{n!}\right)^{\frac{1}{n}}>\frac{1}{2\pi}-\frac{1}{4\pi}=\frac{1}{4\pi}.\ \ \ (1)$$
Of course, 
$$\limsup_{n\rightarrow +\infty}{(n!)}^{\frac{1}{n}}=+\infty$$ 
and we can see that, by observing that $\lim_{n\rightarrow +\infty}(n+1)!/n!=\lim_{n\rightarrow +\infty}(n+1)=+\infty$ (there is a known result, otherwise use AM-GM with that). Therefore, for any $M>0,$ there exists $n_1\in \mathbb{N}^*$ such that $\forall n\geq n_1$ we have that
$${(n!)}^{\frac{1}{n}}>4\pi M.\ \ \ (2)$$
For the arbitrarily chosen $M>0$ the inequalities $(1)$ and $(2)$ imply that
$${|B_n|}^{\frac{1}{n}}=\left(\frac{|B_n|}{n!}\right)^{\frac{1}{n}}\cdot{(n!)}^{\frac{1}{n}}>M,\ \ \ \forall n\geq m.$$
This implies the desired result.
A: The radius of convergence of $f(z)$ is $2\pi$ and so, by the ratio test,
$$\limsup\sqrt[n]{\frac{|B_n|}{n!}}=\frac1{2\pi}.$$
But $\sqrt[n]{n!}\to\infty$ as $n\to\infty$ (by Stirling, or by crude estimates
like $n!\ge(n/2)^{n/2-1}$). Therefore $\limsup\sqrt[n]{|B_n|}=\infty$.
