 Solution

Suppose, towards a contradiction, that $|a_n{z}^n|\le M$ for every $n$, where $|z|>R$. Choose a radius $\rho$ strictly between $R$ and $|z|$, and take any $w$ with $|w|=\rho$. ThenThe 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 $|a_n{z}^n|$ 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:For the first, consecutive absolute terms have ratio $(n+1)|z|$, so at every $z\ne 0$ 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 $|z|/(n+1)\to 0$, 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.