# Convergence almost surely and SLLN

I'm studying probability theory, especially about limit theorem. And I got some trouble in moving forward.

The problem is:

Let $$(X_n)$$ be i.i.d random variables with $$E(X_1)=0$$ and $$E(|X_1|^p)<1$$ for some $$1. Show that $$\displaystyle \frac{S_n}{n^{1/p}}$$ converges to $$0$$ almost surely, where $$S_k=X_1+...+X_k$$.

At first, I tried with Strong Laws of Large Numbers, but it doesn't help. I think the condition that $$E(|X_1|^p)<1$$ for some $$1 is crucial, but I don't figure out what it means.

First note that $$\frac{S_{n}}{n^{\frac{1}{p}}}=\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}$$ Now, since that $$(X_{n})$$ are identically distributed, and since that $$\mathbb{E}[|X_{1}-0|^{p}]<1<\infty \implies \text{absolutely integrable:} \quad \mathbb{E}[X]<\infty$$ So, by The Strong Law of Large Number (SLLN), we can conclude that $$\frac{S_{n}}{n}\overset{a.s}{\to}\mathbb{E}[X]=0$$ So, $$\frac{S_{n}}{\sqrt[p]{n}}=\frac{S_{n}}{n}\cdot \frac{n}{\sqrt[p]{n}}\overset{a.s}{\to} 0\cdot c =0$$
• In the last line, how can $\frac{n}{\sqrt[p]{n}} \to$ a constant almost surely ? Dec 14, 2020 at 23:59