Q:

Showing that the radius of convergence of $\sum_{\nu = 0}^\infty \frac{B_{\nu}}{\nu !}z^\nu$ is at most $2\pi$

Consider the function $f: B_{2\pi}(0) \to \mathbb{C}$ defined by
$$\begin{align}f(z) := \begin{cases} z/(e^z - 1) & z \neq 0\\
1 & z = 0
\end{cases}\end{align}$$
It is well known, that the Taylor expansion of $f$ around $0$ is given by $$\sum_{\nu = 0}^\infty \frac{B_{\nu}}{\nu !}z^\nu$$ where $B_\nu$ are the Bernoulli numbers of the first kind. I know that the radius of convergence of the Taylor expansion is $2\pi$. First, from the expansion lemma we get that the radius of convergence is at least $2\pi$ since $f$ is a holomorphic function in $B_{2\pi}(0)$ and has therefore a Taylor expansion there which converges there. And here is my question: Why is the radius of convergence at most $2\pi$? I mean , I know that $$\lim_{z \to 2\pi i} f(z) = \infty$$ and so also $$\lim_{z \to 2\pi i}\sum_{\nu = 0}^\infty \frac{B_{\nu}}{\nu !}z^\nu = \infty$$ So $\sum_{\nu = 0}^\infty \frac{B_{\nu}}{\nu !}z^\nu$ simply does not converge outside of the disc. Can someone help me formalize my thoughts?

A:

The reason the radius of convergence cannot exceed $2\pi$ is that there are poles at $z=\pm2\pi i.$ At those points, the denominator, $e^z-1,$ is $0$, but the numerator is a non-zero number.

