# Probability ratio convergence in limit

The following is the question :

Given random variables $X_1 \leq X_2 \leq \cdots$ such that $E[X_n] \sim A n^{\alpha}$, where $A,\alpha > 0$ ($\sim$ means that the ratio of the two quantities goes to $1$ as $n$ tends to infinitely). Given that $Var(X_n) \leq Bn^\beta$, where $\beta < 2\alpha$, and $B$ is some constant, please show that convergence of $\frac{X_n}{n^\alpha} \to A$ almost surely.

Note that $X_n$ are not given independent.

What I can conclude is the following : Suppose we drop the asymptotic and just assume equality (we can always compensate for this by fixing some small $\delta > 0$ and choosing sufficiently large $n$ for which the ratio is in $(1-\delta,1+\delta)$) If $Y_n = \frac{X_n}{n^{\alpha}}$, then $\Pr(|Y-A| > \epsilon) \leq \frac{Bn^{\beta-2\alpha}}{\epsilon^2}$, which goes to zero for all $\epsilon$ as $\beta < 2 \alpha$. Hence, we have convergence in probability.

This means, furthermore, that there is an a.s. convergent subsequence $Y_{n_k}$ to $A$. In fact, any subsequence of $Y_n$ will have a subsequence convergent to $A$.

Now, note that $X_n$ are monotonic, but $Y_n$ need not be. So I cannot conclude from the above that $Y_n \to A$. I need help to finish this argument.