Bernoulli Numbers and radius of convergence

Question

consider the function $f(x)=\frac{x}{e^x-1}$.

Since the function $\frac{1}{f(x)}=\frac{e^x-1}{x}=\sum\limits_{k=0}^{\infty} \frac{x^k}{(k+1)!}$ has a taylor expansion with $\frac{1}{f(0)}\neq 0$ we know that $f(x)$ must have a taylor expansion in some neighborhood of 0, i.e. we can write

$f(x)=\sum\limits_{k=0}^{\infty} B_n \frac{x^n}{n!}$ in some neighborhood $(-\delta, \delta)$ with $\delta>0$.

In Wikipedia and other sources, I've read the radius of convergence of the series $\sum\limits_{k=0}^{\infty} B_n \frac{x^n}{n!}$ is actually $(-2\pi,2\pi)$. Is that true and how con one prove it? Further, does the Taylor series converge for all $\vert x \vert< 2\pi$ against f(x) ? How can I see it?

I hope someone can explain it to me.

Best regards

Answer 1

Although it is possible to get distracted by connections to subtler things, observe that $f(z)=z/(e^z-1)$ is holomorphic on $|z|<2\pi$, has simple poles at $\pm 2\pi i$, and is holomorphic on $\{|z|<4\pi\}$ with $\pm 2\pi i$ removed. A power series converges on the largest disk in which a function is holomorphic, so the radius of convergence is $2\pi$.

Answer 2

Try computing the radius of convergence $r = \limsup \sqrt[n]{B_n}$. It is related to the Riemann Zeta function.

$$ \frac{B_n}{2n!} = \zeta(2n)\frac{2 }{(2\pi)^{2n}}$$

with $\zeta(2n) \to 1$ and $2^{1/n} \to 1$.

I wonder then, how to write the analytic continuation, since $f(x)$ is clearly defined outside of $|x| < 2\pi$.