# Power series is locally normally convergent in its convergence radius $B(0,R)$

$$\sum_0^\infty a_nz^n$$ , $$z\in \mathbb{C}$$, a power series with $$R:=\sup \{ t\ge0 : a_n t^n$$ is bounded $$\}$$ as its convergence radius. I wish to prove that $$\sum_0^\infty a_nz^n$$ is locally normally convergent over $$B(0,R)$$.

What I did so far:

• Let $$r_1 , then for all $$z\in B(0,r_1)$$ : $$\sum_0^\infty|a_nz^n| = \sum _0^\infty|a_n|\frac{r_2^n}{r_2^n}|z|^n \le \sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n .$$
• Now, because $$r_2 , $$|a_n|r_2^n , so: $$\sum _0^\infty|a_n|r_2^n(\frac{r_1}{r_2})^n \le M\sum _0^\infty(\frac{r_1}{r_2})^n ,$$ which converges as a geometric series when $$r_1. So any power series for $$z\in B(0,r_1)$$ is mutually bounded by $$M\sum_0^\infty(\frac{r_1}{r_2})^n$$.

Yet I don't succeed to make the next step, and show that any $$z\in B(0,r_1)$$ has a neighborhood $$U_z$$ such that $$\sum_0^\infty \sup_{U_z} |a_n z^n|$$ converges.

• @MartinR That's the point, I really feel its true but I do't succeed to formalize it, I tried to show that the $sup$ series is also bounded by the same geometric series but couldn't justify it... – dan Dec 30 '18 at 17:36
• Actually what I said is nonsense. You need to show that $\sum \sup_U |a_n z^n|$ converges, not $\sum \sup_U |f|$. – Martin R Dec 30 '18 at 17:41
• @MartinR ,Right! I'm sorry I miss typed, just corrected it. – dan Dec 30 '18 at 17:43
• Or in short, for any $z_0\in B(0,r_1)$, $U_{z_0}=B(0,r_1)$ is such an environment. – LutzL Dec 31 '18 at 12:37

For a given $$z_0 \in B(0, R)$$ choose $$r_1, r_2$$ with $$|z_0| < r_1 < r_2 < R$$. Then $$U=B(0, r_1)$$ is a neighbourhood of $$z_0$$, and $$\sum_{n=0}^\infty \sup_U |a_n z^n|$$ is convergent because $$\sup_U |a_n z^n| \le M \left( \frac{r_1}{r_2} \right)^n$$ for some $$M > 0$$.