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Every $R\in [0,1]$ occurs as the radius of convergence of a power series all of whose coefficients are positive integers. For $0<R\le 1$, use the ceiling function to choose $a_n=\lceil R^{-n}\rceil$. ThenFor $|z|<R$, the comparison test for series with a geometric series proves absolute convergence. For $|z|\ge R$, the bound $|a_n{z}^n|\ge (|z|/R{)}^n\ge 1$ violates the fact that terms of a convergent series tend to zero. Hence the radius of convergence is exactly $R$.

For $R=0$, use $a_n=n!$: at every $z\ne 0$ the ratio of successive term absolute values is $(n+1)|z|$, eventually exceeding two. The factorial coefficients therefore give divergence away from zero.

A power series with every coefficient a positive integer cannot have radius of convergence greater than one: $a_n\ge 1$ makes convergence impossible at $z=1$. If zero coefficients are allowed and only finitely many are nonzero, a polynomial instead has infinite radius of convergence.

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.

The comparison test for series states that, for $0\le u_n\le v_n$ eventually, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of $\sum u_n$ implies divergence of $\sum v_n$.

For $|z|<1$, the coefficient bound givesby the geometric series formula and the comparison test for series. The single term $a_0$ does not affect convergence. For $|z|\ge 1$,so the terms do not tend to zero. The fact that terms of a convergent series tend to zero excludes convergence there. Thus

Here $0<\ln 2<1$, so $(\ln 2{)}^{-n}$ grows exponentially rather than decaying. The positive terms $a_n=n^3/(\ln 2{)}^n$ do not tend to zero; indeed they tend to $+\infty$. Therefore the series diverges by the necessary condition that the terms of a convergent series tend to zero.

The ratio test gives the same conclusion: $a_{n+1}/a_n=((n+1)/n{)}^3/\ln 2\to 1/\ln 2>1$.

There is an indexing defect in the original PDF: the term at $n=1$ is undefined because $\ln 1=0$. Thus the series as literally indexed is not defined. The natural finite-index repair is to start at $n=2$, and that repaired series still diverges.

Indeed, for $n\ge 2$ the terms are positive andThey eventually increase at a geometric rate and do not tend to zero. Since the terms of a convergent series tend to zero, no choice of a finite replacement for the missing first term could make the tail converge.