Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 12E / Solution

First establish the interior convergence lemma for a power series. If $\sum a_n{w}^n$ converges at some $w\ne 0$, its terms are bounded: $|a_n{w}^n|\le M$ for a finite $M$. For every $|z|<|w|$,Comparison with a convergent geometric series proves absolute convergence at $z$.

Now letThe set contains zero, since at $w=0$ only the constant term remains. If $|z|<R$, the definition of supremum gives a point of convergence $w$ with $|w|>|z|$. The lemma proves absolute convergence at $z$. If $|z|>R$, convergence would place $|z|$ in the set defining $R$, a contradiction. This proves the existence of a radius of convergence, including $R=0$ and $R=\infty$:The proof leaves behavior on $|z|=R$ open; it need not be uniform around that circle. The two examples illustrate opposite possibilities.