= Solution
Let $M$ bound all $|a_nw^n|$ and put $q=|z|/|w|<1$. Then
$$
|a_nz^n|=|a_nw^n|q^n\le Mq^n.
$$
The <geometric series> $\sum_{n=0}^\infty Mq^n$ converges, so the <comparison test for series> proves \b[absolute convergence] of $\sum a_nz^n$. This is the <bounded power-series terms force interior absolute convergence> principle.
Define the <radius of convergence> by
$$
\boxed{\rho=\sup\{|w|:\ \sum_{n=0}^\infty a_nw^n\text{ converges}\},}
$$
allowing $\rho=0$ and $\rho=\infty$. If $|z|<\rho$, the supremum supplies a point of convergence $w$ with $|w|>|z|$; convergence makes its terms bounded, so the argument above gives <absolute convergence> at $z$. If $|z|>\rho$, convergence is impossible by definition. At $|z|=\rho$, no general conclusion follows. Thus the convergence region has a circular interior centered at zero, with boundary behavior depending on the <power series>.
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