Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 1 / 12F (Analysis I)

Let $L={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. The Cauchy-Hadamard theorem, proved by applying the root test, gives a radius of convergencewith the usual conventions. For $|z|<R$, choose $q<1$ eventually bounding $|a_n{z}^n{|}^{1/n}$, which gives absolute convergence by comparison with a geometric series. For $|z|>R$, infinitely many terms have $n$th root greater than one, so the terms fail to tend to zero and the series diverges.

Here $R=2$ and the assumed limit gives $|a_n{|}^{1/n}\to 1/2$. ThereforeThus

The series can be written asThe original power series converges when $|z{|}^k<2$ and diverges when $|z{|}^k>2$. Hence its radius as a power series in $z$ is

Apply the root test to the terms $a_n{z}^{n^2}$:This tends to zero when $|z|<1$ and to infinity when $|z|>1$. On $|z|=1$, the series converges absolutely because $\sum |a_n|$ converges: the point $1$ lies strictly inside the original radius two. Thus the lacunary power series has radiusand in fact converges on its entire boundary circle.

Set $w=e^z$. ThenThe hypothesis that convergence and divergence both occur implies that this ordinary power series has a radius $R$ with $0<R<\infty$. Since $|w|=e^{\mathrm{Re}z}$, the half-plane of convergence of an exponential power series is determined bythe series converges for $\mathrm{Re}z<S$ and diverges for $\mathrm{Re}z>S$.

For the specified series,Writing $z=x+iy$ gives $|2e^{iz}|=2e^{-y}$. The series converges absolutely when $y>\log 2$ and diverges by the term test when $y<\log 2$. On $y=\log 2$, its absolute values are $(n+1{)}^{-2}$, so it still converges absolutely. Therefore the exact convergence set is