Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 7 / c

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.