= Solution
First establish a comparison fact for a <power series>. If it converges at $w\ne0$, its terms $a_nw^n$ are bounded, say $|a_nw^n|\leq C$. For $|z|<|w|$,
$$
\sum_{n=0}^\infty|a_nz^n|\leq C\sum_{n=0}^\infty\left(\frac{|z|}{|w|}\right)^n<\infty
$$
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 $R$. Therefore
$$
\boxed{\text{convergence for }|z|<R,\qquad\text{divergence for }|z|>R.}
$$
The 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 $n\geq1$, 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. Hence
$$
\boxed{R=1.}
$$
For instance the positive boundary point gives the divergent <harmonic series>; no assertion of universal boundary convergence is needed.
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