Let's say $A_n=O_p(n^{-3/5}\sqrt{log n})$. I wonder whether I can express $A_n$ as $o_p(n^{-2/5})$.Here's my attempt.
$$ P(\frac{|A_n|}{|n^{-3/5}\sqrt{log n}|} > \lambda) < \varepsilon $$ for all $n > M$ for some $M >0$. This implies that
$$P(|A_n| > \lambda \cdot n^{-2/5}n^{-1/5}\sqrt{log n}) < \varepsilon .$$
Since $lim_{n \rightarrow \infty} \frac{\sqrt{log n}}{n^{1/5}}=0$ by L'Hopital's rule, I can express the probability inequality as
$$P(\frac{|A_n|}{n^{-2/5}}> \varepsilon \prime) \rightarrow 0 .$$ Thus,
$$A_n=o_p(n^{-2/5}).$$ Am I right ?
