= Solution
A complex <power series> with <radius of convergence> $R$ converges absolutely at every $z$ with $|z|<R$ and diverges at every $z$ with $|z|>R$. This makes no general assertion at $|z|=R$: boundary points must be examined separately. The series always converges at $z=0$. If $R=0$, no nonzero point lies in the convergence disk; if $R=\infty$, the series converges absolutely throughout the <complex plane>.
If $p$ is the zero polynomial, every coefficient vanishes, so \b[$R=\infty$]. Otherwise let $d$ be its degree and $c\ne0$ its leading coefficient. Since $p(n)=cn^d(1+o(1))$,
$$
\left|\frac{p(n+1)}{p(n)}\right|\longrightarrow1.
$$
The <ratio test> therefore proves absolute convergence when $|z|<1$. For $|z|>1$, the terms $p(n)z^n$ do not tend to zero, so the series diverges. Thus the <polynomial-coefficient power series> has \b[$R=1$ for every nonzero polynomial]. In fact, it also diverges at every $|z|=1$, because $|p(n)z^n|=|p(n)|$ does not tend to zero, whether $d=0$ or $d>0$.
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