absolut and uniform convergence of complex power series on an open disk imply absolut and local uniform convergence on the whole disk of convergence

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I'm stuck in the proof of the following theorem of complex analysis:

There is a $R\in\mathbb{R}_{0}^{+}\cup\{\infty\}$ s.t.:

(i) $\sum_{k=0}^{\infty} a_k (z-z_0)^k$ converges absolutely and locally uniformly (i.e. every point has a ngbh. in which the convergence is uniformly) on $D_R(z_0)=\{z\in\mathbb{C}\mid\lvert z-z_0\rvert< R\}$

(ii) The power series diverges on $\mathbb{C}\backslash\{\overline{D_{R}(z_0)}\}$. The Cauchy-Hadamard Theorem holds: $$R=\frac{1}{\limsup\limits_{n\rightarrow\infty}\sqrt[n]{\lvert a_n\rvert}}$$

The proof starts as follows:

General idea: majorizing by the Geometric Series

Let $0<r_0<r_1<R$. If $R=0$ consider (ii). We want to show absolut and uniform convergence on $D_{r_0}(z_0)$; this implies the theorem. [...]

Here I'm stuck. Why does this imply the theorem? I don't have any counterexample but have even less a clue to show this implication. In the proof there is no further explanation given why this implication holds.

Could anyone please proof this implication or refer to some open resource? Many thanks in advance!