The following is a execise from Durrett's: Probability: Theory and Examples.
Let $p>0$. Let $X_i$ be i.i.d random variables such that $EX_i =0$, and define $S_n = \sum_{i=1}^n X_i$. Show that if $\dfrac{S_n}{n^{1/p}}\to 0\,$ almost surely, then $E|X_i|^p < \infty$.
He says it's an easy exercise so I imagine there is a simple trick here.
To start, I know that
$$E|X_i|^p = \int_0^\infty P(|X_i|^p > x)dx = \int_0^\infty P(|X_i| > x^{1/p}) dx \leq 1+ \sum_{n=1}^\infty P(|X_n|>n^{1/p})$$
which I think might be helpful. I also think a contrapositive proof might be best?
Suppose $E|X_i|^p = \infty$. Then this implies $\sum_{n=1}^\infty P(|X_n|>n^{1/p}) = \infty$. Since the $X_i$'s are independent, the second Borel-Cantelli lemma then tells us:
$$P(|X_n|>n^{1/p} \quad i.o) = 1.$$
And I think that this implies $S_n/n^{1/p}$ cannot converge to $0$, since for the above implies $S_n/n^{1/p}>1$ infinitely often. Hence, giving the result by contrapositive.
I'm really not confident with my probability theory. So I want to check my reasoning is correct here.
You indeed got the good idea to show the the link between integrability of $\left\lvert X_1\right\rvert^p$ and the fact that $|X_n|>n^{1/p}$ infinitely often almost surely. However, it is not so clear that it implies that $S_n/n^{1/p}>1$, essentially due to the absolute values. However, $$ \frac{X_n}{n^{1/p}}=\frac{S_n}{n^{1/p}}-\frac{S_{n-1}}{(n-1)^{1/p}}\left(\frac{n-1}n\right)^{1/p} $$ hence $S_n/n^{1/p}\to 0$ almost surely implies that $X_n/n^{1/p}\to 0$ a.s., which finishes the argument.