 Solution

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.