= Solution
First establish the interior convergence lemma for a <power series>. If $\sum a_nw^n$ converges at some $w\ne0$, its terms are bounded: $|a_nw^n|\leq M$ for a finite $M$. For every $|z|<|w|$,
$$
|a_nz^n|\leq M\left(\frac{|z|}{|w|}\right)^n.
$$
Comparison with a convergent <geometric series> proves <absolute convergence> at $z$.
Now let
$$
R=\sup\{|w|:\textstyle\sum a_nw^n\text{ converges}\}\in[0,\infty].
$$
The set contains zero, since at $w=0$ only the constant term remains. If $|z|<R$, the definition of <supremum> gives a point of convergence $w$ with $|w|>|z|$. The lemma proves <absolute convergence> at $z$. If $|z|>R$, convergence would place $|z|$ in the set defining $R$, a contradiction. This proves the existence of a <radius of convergence>, including $R=0$ and $R=\infty$:
$$
\boxed{\text{Convergence for }|z|<R,\qquad\text{divergence for }|z|>R.}
$$
The proof leaves behavior on $|z|=R$ open; it need not be uniform around that circle. The two examples illustrate opposite possibilities.
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