Let $\{M_n\}$ be a martingale with bounded increments: $|M_n-M_{n-1}|\leq 1$. $M_0=0$. Prove that $\frac{M_n}{n^{\alpha}}$ converges to $0$ a.s. for $\alpha>\frac{1}{2}$.
For this problem, my solution is as follows. I am not sure whether it is correct or not. I first applied Doob's maximal inequality to the non-negative submartingale $\{|M_n|^2\}$ to get some estimate on the deviation:
\begin{align*} P\{\max_{1\leq i\leq n}{|M_i|}>\epsilon n^\alpha\}&=P\{\max_{1\leq i\leq n}{|M_i|^2}>\epsilon^2 n^{2\alpha}\}\\ &\leq \frac{E[|M_n|^2]}{\epsilon^2 n^{2\alpha}}\\ &=\frac{\sum_{i=1}^nE[(M_i-M_{i-1})^2]}{\epsilon^2 n^{2\alpha}}\\ &\leq\frac{1}{\epsilon^2 n^{2\alpha-1}} \end{align*}
The third equality uses the fact that $L^2$ martingales have orthogonal increments. Second, I took a subsequence $a_n=[n^{\frac{2}{2\alpha-1}}]+1$ and used the above estimate along with Borel-Cantelli lemma to the subsequence $\{M_{a_n}\}$. This gives that for almost every $\omega\in \Omega$, there exists an integer $N(\omega)$ such that
$$ \frac{|M_{a_n}|}{a_n^\alpha}\leq\frac{\max_{1\leq i\leq a_n}{|M_i|}}{a_n^\alpha}\leq\epsilon,\ \ \forall n>N(\omega) $$
Particularly, this proves the desired result for the subsequence $\{M_{a_n}\}$. Finally I extended the result to the whole sequence using the sandwiching technique. For each $n$, there exists a unique $k$ such that $a_{k}\leq n<a_{k+1}$. Therefore,
\begin{align*} \frac{|M_{n}|}{n^\alpha}&\leq \frac{\max_{1\leq i\leq a_{k+1}}{|M_i|}}{a_{k+1}^\alpha}\frac{a_{k+1}^\alpha}{a_{k}^\alpha}\\ &=\frac{\max_{1\leq i\leq a_{k+1}}{|M_i|}}{a_{k+1}^\alpha}(\frac{k+1}{k})^{\frac{2\alpha}{2\alpha-1}} \end{align*}
The last equality is actually not equality but I write equality for convenience. Tending $n\rightarrow\infty$, then accordingly $k\rightarrow \infty$, the right-hand side goes to $0$. Hence we conclude that $\frac{M_n}{n^{\alpha}}$ converges to $0$ a.s. for $\alpha>\frac{1}{2}$.