= Solution
The lacunary series converges locally uniformly on $|z|<1$: for $|z|\le r<1$, $\sum r^{2^n}\le\sum_{j\ge1}r^j<\infty$. 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 $r\uparrow1$, the sum of any fixed number of these tail terms tends to that number, while the earlier finite sum stays bounded. Thus $\operatorname{Re}f(r\zeta)\to+\infty$. 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 \b[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.
Back to article page
