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A power series has radius of convergence $R$ when it converges absolutely for every $|z|<R$ and diverges for every $|z|>R$. Its behaviour on $|z|=R$ is not specified by this definition. When $R=0$, it converges at the centre and diverges at every nonzero point; an infinite radius of convergence means absolute convergence at every finite $z$.

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for all $n$ at some $|z|>R$. Choose a complex number $w$ with $R<|w|<|z|$. ThenThe comparison test for series with the convergent geometric series $\sum M{q}^n$ yields absolute convergence at $w$, contradicting the definition of $R$. Thus terms of a power series are unbounded outside its convergence disc:For the given power series, the ratio test gives, when $z\ne 0$,Hence it converges absolutely for $|z|<1$ and its terms fail to tend to zero for $|z|>1$ (the ratios are eventually larger than some number greater than $1$). Therefore $\boxed{R=1}$. In fact, on $|z|=1$ it also converges absolutely by comparison with the convergent p-series $\sum n^{-3}$.