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
Best Answer
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 $2\pi$.