I encountered this problem from Durrett (Exercise 2.5.1):
Suppose $X_1, X_2, \dots$ are i.i.d. with $EX_i = 0$, $var(X_i) = C< \infty$. Use Kolmogorov's maximal inequality to prove that if $S_n = \sum_{i=1}^nX_i$ and $p > 1/2$, then $S_n/n^p \to 0$ a.s.
The problem says to prove it using Kolmogorov's maximal inequality, but I did not see how one would use it and I cannot really understand the solution proof to Durrett. So I came up with my own proof and I want to see if my proof is correct.
My Proof: Pick $\alpha$ such that $\alpha(2p-1) > 1$. Let the subsequence $n(m) = [m^\alpha]+1$ where $[m^\alpha]$ is the integer part of $m^\alpha$. Fix $\epsilon > 0$, then
$$
P(|S_{n(m)}| > \epsilon n(m)^p) \leq \frac{ES_{n(m)}^2}{\epsilon^2n(m)^{2p}} = \frac{C}{\epsilon^2n(m)^{2p-1}} \leq \frac{C}{\epsilon^2m^{\alpha(2p-1)}}.
$$
Since the RHS of the above is summable, by BC lemma, $S_{n(m)}/n(m)^p \to 0$ a.s.
Now since for any sequence $x_n$
$$
\frac{x_n}{n^p} \to 0 \Longleftrightarrow \frac{\max_{m\leq n}|x_m|}{n^p} \to 0,
$$
we have that
$$
\frac{\max_{k \leq n(m)}|S_{k}|}{n(m)^p} \to 0 \ a.s.
$$
Now since $\max_{k \leq n}|S_k|$ is increasing in $n$, then for $n(m) \leq n \leq n(m+1)$, we have
$$
\frac{\max_{k \leq n(m)}|S_k|}{n(m+1)^p} \leq \frac{\max_{k \leq n}|S_{k}|}{n^p} \leq \frac{\max_{k \leq n(m+1)}|S_{k}|}{n(m)^p}.
$$
Then we have
$$
\frac{\max_{k \leq n(m)}|S_k|}{n(m)^p}\frac{n(m)^p}{n(m+1)^p} \leq \frac{\max_{k \leq n}|S_{k}|}{n^p} \leq \frac{\max_{k \leq n(m+1)}|S_{k}|}{n(m+1)^p}\frac{n(m+1)^p}{n(m)^p}.
$$
Now since
$$
\frac{n(m)^p}{n(m+1)^p} = \left(\frac{[m^\alpha]+1}{[(m+1)^\alpha]+1}\right)^p \to 1
$$
and
$$
\frac{n(m+1)^p}{n(m)^p} = \left(\frac{[(m+1)^\alpha]+1}{[m^\alpha]+1}\right)^p\to 1,
$$
we can then conclude that
$$
\frac{\max_{k\leq n}|S_k|}{n^p} \to 0 \ a.s.
$$
and
$$
\frac{S_n}{n^p} \to 0 \ a.s.
$$
Is my proof correct? Thank you!
Proof Correction:
As per comment of Davide, the proof should correct the start of the argument and use Kolmogorov's maximal inequality:
$$
P(\max_{k\leq n(m)}|S_k| > \epsilon n(m)^p) \leq \frac{ES_{n(m)}^p}{\epsilon^2n(m)^{2p}} \leq \frac{C}{\epsilon^2m^{\alpha(2p-1)}}.
$$
Hence,
$$
\frac{\max_{k\leq n(m)}|S_k|}{n(m)^p} \to 0 \ a.s.
$$
The rest of the proof should follow in the same as my original proof.
Best Answer
It seems that there is a gap in your proof: you proved that $S_{n(m)}/n(m)^p \to 0$ almost surely and the deterministic fact you mention, namely, $\frac{x_n}{n^p} \to 0 \Longleftrightarrow \frac{\max_{m\leq n}|x_m|}{n^p} \to 0$, would only allow to deduce that $\max_{1\leqslant k\leqslant m}\lvert S_{n(k)}\rvert/n(m)^p\to 0$ in probability. In other words, this gives no information about $S_j$ when $j$ is not of the form $n(k)$ for some integer $k$.
Nevertheless, the proof can be saved by using Kolmogorov's maximal inequality in order to bound $\mathbb P\left(\max_{1\leqslant k\leqslant m(n)}|S_{k}| > \epsilon n(m)^p\right)$ by a similar term as the one obtained in the opening post.