= Solution
Suppose, towards a contradiction, that $|a_nz^n|\le M$ for every $n$, where $|z|>R$. Choose a radius $\rho$ strictly between $R$ and $|z|$, and take any $w$ with $|w|=\rho$. Then
$$
|a_nw^n|=|a_nz^n|\left|\frac wz\right|^n\le M\left(\frac\rho{|z|}\right)^n.
$$
The <geometric series> on the right converges because its ratio lies in $[0,1)$. The <comparison test for series> therefore gives <absolute convergence> of the <power series> at $w$, contrary to the defining property of its <radius of convergence>. Thus \b[$|a_nz^n|$ is unbounded in $n$ whenever $|z|>R$]. This is the <bounded power-series terms force interior absolute convergence> principle. The case $R=\infty$ has no points outside its convergence disk and the assertion is vacuous there.
Every possible <radius of convergence> is realized by an explicit <power series>:
$$
\boxed{\begin{array}{c|c}R&\text{series}\\\hline 0&\displaystyle\sum_{n=0}^\infty n!z^n\\0<R<\infty&\displaystyle\sum_{n=0}^\infty(z/R)^n\\\infty&\displaystyle\sum_{n=0}^\infty z^n/n!\end{array}}
$$
For the first, consecutive absolute terms have ratio $(n+1)|z|$, so at every $z\ne0$ they eventually grow and do not tend to zero; at zero the series is defined. The middle <geometric series> converges absolutely exactly for $|z|<R$, and outside that disk its terms grow. The last has consecutive-term ratio $|z|/(n+1)\to0$, so the <ratio test> gives <absolute convergence> for every finite $z$. These arguments use the necessary zero-term condition for convergence, the <comparison test for series>, and <absolute convergence> of a <geometric series> with ratio less than one; they do not assume convergence on the boundary.
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