Showing $\frac{S_n}{n^{1/p+\varepsilon}}$ converges to $0$ a.e.

Question

Let $\{X_n\}$ be independent and identically distributed random variables such that $E(|X_1|^p) < \infty $ for some $p$ satisfying $0 < p < 2$; in case $p > 1$, we assume that $E(X_1) = 0$, then prove that $S_n n^{-(1/p)-\varepsilon}$ converges to $0$ a.e.

Approach:

Since $E(|X_{1}|^{p}) < \infty$ , it implies $\frac{S_n}{n}$ converges to $E(|X_{1}|)$ a.e. And $E(|X_{1}|)=0$ is given.

That means $\frac{S_n}{n}$ converges to $0$ a.e. How do I bring in the factor $n^{−1/p−\varepsilon}$ ?

I am stuck after that. What should I start with?