= Solution
Convergence at $z_0$ implies that the terms $a_nz_0^n$ form a <bounded sequence>: $|a_nz_0^n|\leq M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, then
$$
|a_nz^n|=|a_nz_0^n|r^n\leq Mr^n.
$$
Comparison with the convergent <geometric series> $\sum Mr^n$ proves <absolute convergence>. The case $z_0=0$ is vacuous.
The <radius of convergence> is the number $R\in[0,\infty]$ such that a <power series> converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.
Take
$$
v=-1,\qquad w=1.
$$
Then $\sum v^n/n$ is the convergent <alternating harmonic series>, whereas $\sum w^n/n$ is the divergent <harmonic series>. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that
$$
\boxed{R=1}.
$$
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