Question: If $X_n$ is any sequence of random variables, there are constants $c_n\to\infty$ so that $X_n/c_n\to 0$ a.s.
I know "$X_n\to 0$ iff for all $\epsilon >0,\space P(|X_n|>\epsilon\space \text{i.o.})=0$", and for that I have to use Borel-Cantelli Lemma.
By Chebychev's inequality, for any $\epsilon>0$, $P(|X_n|> |c_n|\epsilon)\leq \dfrac{EX^2_n}{c^2_n\epsilon^2}$. Let, $EX^2_n=O(n^k)$, then I choose $|c_n|=O(n^{\frac{k}{2}+1})$. In that case, $P(|X_n|> |c_n|\epsilon)\leq\dfrac{\delta}{n^2}$, for some positive constant $\delta$.
Since, $$\sum_{n=1}^{\infty}P\Big(\Big|\dfrac{X_n}{c_n}\Big|>\epsilon\Big)<\infty$$
then, by Borel-Cantelli Lemma, $P\Big(\Big|\dfrac{X_n}{c_n}\Big|>\epsilon\space\text{i.o.}\Big)=0$
Hence, $X_n/c_n\to 0$ a.s.
Is my argument ok? Thanks for your help.
Note: i.o. means infinitely often.