$X_n$ is an increasing sequence of non-negative random variables with $\mathbb{E}(X_n)\sim an^\alpha$ and $\text{Var}(X_n)\sim bn^\beta$ with $a,b,\alpha>0$ and $\beta<2\alpha$. Show that $\dfrac{X_n}{n^\alpha}\to a \;\;\text{a.e}$.
Now I am having trouble on how to use the fact that the means are approximately of the form given, and not exactly equal to it. Also the increasing nature must be required somewhere that I can't think of. Can someone give me some hints on how to proceed? Thanks a lot.
You will need to first use convergence over a sparse subsequence $n[k]$. So, first use the Markov/Chebyshev inequality to prove (with prob 1):
$$\lim_{k\rightarrow\infty}\frac{X_{n[k]}-E[X_{n[k]}]}{n[k]^{\alpha}}=0$$
You can use, for example, $n[k]=k^r$ for a suitably large positive integer $r$, or $n[k]=\lceil(1+\delta)^k\rceil$ for any $\delta>0$. Once this is established, you can use the other given properties of $X_n$ to prove the desired convergence on the actual sequence.
When they say $E[X_n] \sim an^{\alpha}$, I think they mean: $$ \lim_{n\rightarrow\infty} \frac{E[X_n]}{an^{\alpha}} = 1 $$ and so it is okay that they do not give you the exact mean, this limiting property is sufficient.