Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 22G / i / Solution

The lacunary series converges locally uniformly on $|z|<1$: for $|z|\le r<1$, $\sum r^{2^n}\le {\sum }_{j\ge 1}{r}^j<\infty$. It therefore defines a holomorphic function there. Let ${\zeta }^{2^N}=1$. For $n\ge N$, $(r\zeta {)}^{2^n}=r^{2^n}$ is positive real. As $r\uparrow 1$, the sum of any fixed number of these tail terms tends to that number, while the earlier finite sum stays bounded. Thus $\mathrm{Re}f(r\zeta )\to +\infty$. A analytic continuation through $\zeta$ would be bounded in a small closed neighborhood and cannot have this radial divergence.

Roots of unity of orders $2^N$ are dense on the unit circle. If an analytic continuation existed through any boundary point, its neighborhood would contain one of these singular roots, and the identity theorem would make it agree with $f$ on the inside portion. This contradicts the divergence just proved. Hence the entire unit circle is a natural boundary. The exponent printed in the PDF is $2^n$, not $2n$; the latter would give a rational function and would not have this property.