 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 $R=\infty$. Otherwise let $d$ be its degree and $c\ne 0$ its leading coefficient. Since $p(n)=c{n}^d(1+o(1))$,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 $R=1$ 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 $d>0$.