Geometric-mean refinement of an asymptotic sequence

For ${\chi }_0={\varphi }_0$ and ${\chi }_n=\sqrt{{\varphi }_{n-1}{\varphi }_n}$, the sequence $({\chi }_n)$ is asymptotic, but ${\chi }_n/{\varphi }_n\to \infty$ for every $n\ge 1$.