= Solution
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 $|a_nz^n|\leq M$ for all $n$ at some $|z|>R$. Choose a <complex number> $w$ with $R<|w|<|z|$. Then
$$
|a_nw^n|=|a_nz^n|\left(\frac{|w|}{|z|}\right)^n
\leq Mq^n,\qquad q=\frac{|w|}{|z|}<1.
$$
The <comparison test for series> with the convergent <geometric series> $\sum Mq^n$ yields <absolute convergence> at $w$, contradicting the definition of $R$. Thus <terms of a power series are unbounded outside its convergence disc>:
$$
\boxed{|z|>R\ \Longrightarrow\ (|a_nz^n|)_{n\geq0}\text{ is unbounded}.}
$$
For the given <power series>, the <ratio test> gives, when $z\ne0$,
$$
\frac{|z|^{n+1}/(n+1)^3}{|z|^n/n^3}
=|z|\left(\frac n{n+1}\right)^3\longrightarrow |z|.
$$
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> $\sum n^{-3}$.
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