= Solution
Let
$$
S=\left\{r\geq0:\sum_{n=1}^\infty |a_n|r^n<\infty\right\}.
$$
The set $S$ contains zero. If $r\in S$ and $0\leq s<r$, then $s\in S$ by comparison. Moreover, if the original <power series> diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.
If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_nx^n$ would be bounded. For any $y$ with $R<|y|<|x|$,
$$
|a_ny^n|\leq M\left|\frac yx\right|^n,
$$
so the <geometric series> comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the <radius of convergence> property.
For $\sum_{n\geq1}x^n/3^n$, the ratio is $x/3$, so the <geometric series> criterion gives
$$
\boxed{R=3}.
$$
Back to article page
