Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 32E / a

For $n\ge 1$, put $r_n={\varphi }_n/{\varphi }_{n-1}$. Since $({\varphi }_n)$ is an asymptotic sequence, $r_n\to 0$. We can rewritesoTogether with ${\psi }_0={\varphi }_0$, this provesMoreover,Hence $({\psi }_n)$ is an asymptotic sequence. This is the harmonic-mean refinement of an asymptotic sequence.

First,For $n\ge 1$,Thus $({\chi }_n)$ is an asymptotic sequence.

Termwise asymptotic equivalence fails for every $n\ge 1$, becauseFor example, as $x\to 0$, the positive scale ${\varphi }_n(x)=|x{|}^n$ gives ${\chi }_n(x)=|x{|}^{n-1/2}$ for $n\ge 1$. This illustrates the geometric-mean refinement of an asymptotic sequence.