Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 1 / 12E (Analysis I)

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.

For $z\ne 0$, the ratio of successive absolute terms isThe ratio test therefore gives absolute convergence for $|z|<1$ and divergence for $|z|>1$, while at zero the series is trivially convergent. Hence $R=1$. On the boundary, $|z^n/n^2|=1/n^2$, and the convergent p-series proves

For $|z|=r<1$, the inequality $n!\ge n$ for $n\ge 1$ givesThe first $r$ comes from $0!=1$; both initial terms have exponent one. Thus the factorial-gap power series converges absolutely inside the unit circle. For $|z|>1$, its terms have magnitudes $|z{|}^{n!}\to \infty$, so the term test for divergence proves divergence and $R=1$.

At every point with $|z|=1$, each term has magnitude one, so the terms cannot tend to zero, regardless of their phases. HenceThe large gaps between exponents make this a lacunary power series; cancellation cannot overcome the necessary condition that the terms of a convergent series tend to zero.