= Solution
The comparison test says that if $0\leq b_n\leq c_n$ eventually and $\sum c_n$ converges, then $\sum b_n$ converges. If $\sum a_nz_0^n$ converges, its terms are bounded: $|a_nz_0^n|\leq M$. For $|z_1|<|z_0|$,
$$
|a_nz_1^n|\leq M\left|\frac{z_1}{z_0}\right|^n,
$$
so comparison with a geometric <series> proves absolute convergence.
The radius $R$ is the number for which the power <series> converges absolutely for $|z|<R$ and diverges for $|z|>R$. Given $r_i<R_i$, both $\sum|a_n|r_1^n$ and $\sum|b_n|r_2^n$ converge, so their terms are bounded, say $|a_n|r_1^n\leq M$. Then
$$
\sum|a_nb_n|(r_1r_2)^n
\leq M\sum|b_n|r_2^n<\infty.
$$
\b[Thus $R\geq r_1r_2$; letting $r_i\uparrow R_i$ gives $R\geq R_1R_2$.]
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