If $Q < \frac{n}{m^2} X_n$ where $X_n$ is a sequence of random variables, $X_n \xrightarrow{a.s}1$, $0\leq Q \leq1$, $m=\omega(\sqrt{n})$ (The $\omega$ denotes the order, see here). Then, how can we conclude $Q\xrightarrow{a.s}0$.
My solution,
Since $P(\lim\limits_{n \to \infty}X_n = 1)=1$ and $Q \frac{m^2}{n}<X_n$,
$$P(\lim\limits_{n \to \infty}Q \frac{m^2}{n}<\lim\limits_{n \to \infty} X_n = 1)=1$$
I know I am almost done, but I don't know how to write it precisely. I'd appreciate it if you could help.
$n/m^2\to 0$, for otherwise if $\liminf_{n\to\infty} n/m^2\to c>0$, then $\liminf_{n\to\infty} \sqrt{n}/m\to c'>0$, so for all large $n, m<\sqrt{n}/(c'-\epsilon)$. so $m=O(\sqrt{n})$, so $m=\Theta(\sqrt{n})$, which contradicts $m=\omega(\sqrt{n})$ according to your link.
$X_n\to 1$ a.s. means outside a set of probability $0$, $(n/m^2)X_n\to (0)(1)=0$, i.e., $(n/m^2)X_n\to 0$ a.s. See also Slutsky's theorem.