$ \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 <r_2<R$ , 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<R$ , $ |a_n|r^n <M$ , 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<r_2$ . 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.