= Solution
The <radius of convergence> of the <power series> $\sum_{n\geq0}a_nz^n$ is
$$
R=\sup\{r\geq0:\text{the series converges for every }|z|<r\}.
$$
If $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_nw^n\to0$, those terms are <bounded sequence>[bounded], say $|a_nw^n|\leq M$. Therefore
$$
|a_nz^n|\leq M\left|\frac zw\right|^n,
$$
and comparison with a <geometric series> proves <absolute convergence>. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.
If $|a_n|\leq|b_n|$, convergence of $\sum b_nw^n$ implies absolute convergence of $\sum a_nz^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that \b[$R_a\geq R_b$].
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