 Solution

First establish a comparison fact for a power series. If it converges at $w\ne 0$, its terms $a_n{w}^n$ are bounded, say $|a_n{w}^n|\le C$. For $|z|<|w|$,by the geometric series. Thus convergence at one point implies absolute convergence at every point of smaller modulus.

Let $S$ be the set of moduli of points at which the power series converges. It contains zero. Define the radius of convergence $R=\sup S\in [0,\infty ]$. If $|z|<R$, the definition of supremum gives a convergent point $w$ with $|w|>|z|$; the comparison just proved gives absolute convergence at $z$. This reasoning also applies when $R=\infty$. If $|z|>R$ and the power series converged there, its modulus would belong to $S$, contradicting the definition of $R$. ThereforeThe boundary $|z|=R$ is intentionally undecided by this argument. If $R=0$, only the exterior assertion is nonvacuous.

For the particular coefficients $a_n=1/n$ for $n\ge 1$, comparison with $\sum |z{|}^n$ gives absolute convergence whenever $|z|<1$. If $|z|=r>1$, then $r^n/n$ fails to tend to zero: its successive ratio is $rn/(n+1)$, eventually greater than a fixed number bigger than one. The necessary term condition for convergence of a series therefore fails. HenceFor instance the positive boundary point gives the divergent harmonic series; no assertion of universal boundary convergence is needed.