How do I prove that the strong law of large numbers is equivalent to $P(∩_{n=1}^∞ ∪_{{k}\geq{n}}{|{\frac{1}{k}}S_k-p|\geq{ε}})=0$

Question

I want to solve the following task:

Show that the strong law of large numbers statement is equivalent to the following statement: For all ε>0: $$P(∩_{n=1}^∞∪_{{k}\geq{n}}{|{\frac{1}{k}}S_k-p|\geq{ε}})=0$$

The strong law of large numbers is: $$P(ω\inΩ: \lim\limits_{n\to\infty}\frac{S_{n}(ω)}{n}=p)=1$$

I just thought about:

1. $$S_n(ω)=\sum \limits_{i=1}^{n}X_i(ω)=X_1(ω)+X_2(ω)+...+X_n(ω)$$

2. $$P(X_i=1)=p$$

3. $${|{\frac{1}{k}}S_k-p|\geq{ε}}$$ $$\Leftrightarrow$$ $$-εk+pk≥S_{k}≥εk+pk$$

I think that's all I need to solve the task. Does anybody have an idea? Thank you:)