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If I consider the following power series, the radius of convergence of each of them is the distance to the nearest singularity in the complex plane: $$e^z=\displaystyle\sum_{n=0}^\infty \dfrac{z^n}{n!}, \, R\to\infty \, \text{(converges everywhere)}$$ $$\ln (1+z)=-\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^n z^n}{n}, \, R=1 \, \text{(diverges at} \, z=-1)$$ $$\arctan z=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n z^{2n+1}}{2n+1}, \, R=1 \, \text{(diverges at} \, z=i)$$ $$\dfrac{z}{e^z-1}=\displaystyle\sum_{n=0}^\infty \dfrac{B_{n}z^n}{n!}, \, R=2\pi \, \text{(diverges at} \, z=2\pi i).$$ Does this hold true for all power series? It's clear that $R\le d$, but can it be shown that $R=d$? The letter $d$ denotes the distance to the nearest singularity in the complex plane (from the origin).

This does not account for removable singularities.

Kaju
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Sure. You are assuming that $f(z)$ is complex analytic on the the closed ball of radius $Q$ for all $Q < d$. Hence $f(z)$ is continuous and thus $|f(z)|$ is bounded by a constant $C$ on the circle $|z| = Q$. But then $f(z) = \sum a_n z^n$ where

$$a_n = \frac{1}{2 \pi i} \oint_{|z| = Q} \frac{f(z)}{z^{n+1}} dz$$

But $|f(z)/z^{n+1}| = |f(z)|/Q^{n+1} \le C/Q^{n+1}$ on the circle $|z| = Q$, so

$$|a_n| \le \frac{1}{2 \pi} \frac{2 \pi C}{Q^{n+1}} = \frac{C}{Q^{n+1}},$$

and so $1/R = \limsup |a_n|^{1/n} \le 1/Q$. Hence $Q \le R$ for all $Q < d$, and so $d \le R$.

Integrand
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user687721
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