## Power series is locally normally converges 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 converges 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 , $$|a_n|r^n , So: $$\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$$, which converge has a geometric series when $$r_1. So any power series $$z\in B(0,r_1)$$ is mutually bounded by $$M\sum_0^\infty\frac{r_1}{r_2}$$.

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 $$\sum_0^\infty \sup{_U{_z}} |f|$$ converges.