Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 4D (AnalysisPI)

Suppose the series converges at $w\ne 0$. Then its terms $a_n{w}^n$ are bounded, say by $M$. Whenever $|z|<|w|$,so comparison with a geometric series proves absolute convergence at $z$.

Letwith the natural values $0$ or $\infty$ allowed. If $|z|<R$, choose a convergent point $w$ with $|z|<|w|\le R$; the preceding argument gives convergence at $z$. If $|z|>R$, convergence at $z$ would contradict the definition of $R$. Hence the series converges for $|z|<R$ and diverges for $|z|>R$. No universal assertion is possible on $|z|=R$.

By the Cauchy-Hadamard theorem, putThe squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_n{z}^n$ and $\sum a_n^2{z}^n$ have radius one.

For the coefficients $a_n^{a_n}$, the relevant root terms areEquality can again occur. For example, takeThen $a_n\to \infty$, $a_n^{1/n}\to 1$, andThus both $\sum a_n{z}^n$ and $\sum a_n^{a_n}{z}^n$ have radius one. These examples illustrate the radius of convergence after powering coefficients.