I have what seems like a simple question, but I am not quite able to prove it. Suppose that $A_n$ is a sequence of random variables, and we know that $A_n$ strongly converges to zero; i.e. almost surely. I am aware that this implies that $A_n$ converges in probability to $0$, but I am wondering about the rate of convergence.
Suppose that I wish to establish that $A_n = \mathcal O_P(f(n))$, where $f(n)$ is some positive, strictly monotonically decreasing function which decays to zero at some rate. My understanding is that strong convergence implies that for any $\epsilon$ there are finitely many $n$ such that $A_n > \epsilon$. But does this condition imply that there is some constant $C_f$, potentially quite large!, such that for $n>>0$, $$\Pr\left(A_n > C_f f(n)\right) \to 0 ?$$