I would like some help on the following problem. Thanks for any help in advance.
Let p > 0. If $S_{n}/n^{1/p} \rightarrow 0$ a.s., then $E|X|^{p}$ < $\infty$.
My attempt so far:
Suppose $E|X|^{p}$ = $\infty$. Let Z = $|X|^{p}$. Then $E(X) \leq \sum_{n=1}^{\infty} P(|Z| \geq n) = \sum_{n=1}^{\infty} P(|X| \geq n^{1/p})$. Then by the Borel-Cantelli lemma $|X| \geq n^{1/p}$ i.o. I'm not sure how to proceed after this.
I suppose $(X_n)$ is i.i.d and $S_n=X_1+X_2+...+X_n$.
$\frac {X_n} {n^{1/p}}=\frac {S_n-S_{n-1}} {n^{1/p}}=\frac {S_n} {n^{1/p}} -\frac {(n-1)^{1/p}} {n^{1/p}} \frac {S_{n-1}} {(n-1)^{1/p}} \to 0$ almost surely. Now Borel Cantelli Lemmas gives $\sum P(|X|>n^{1/p}) <\infty$ and hence $E|X|^{p} <\infty$.