# prove convergence almost surely

Let {$X_n$} be a sequence of i.i.d. random variables with $E[|X_1|]<\infty$, and $S_n=\sum_{j=1}^nX_j$. Show that if $E[X_1]\neq0$, then $$\frac{\max_{1\leq k\leq n} |X_k|}{|S_n|}\rightarrow 0$$ almost surely.

I'm thinking of using strong law of large numbers to prove this, but don't know how to start. Please help.

$$\dfrac{\max_{1\leq k\leq n} |X_k|}{|S_n|} = \dfrac{\dfrac{\max_{1\leq k\leq n} |X_k|}{n}}{\left|\dfrac{S_n}{n}\right|}$$
since $\dfrac{S_n}{n} \to EX_1$ almost surely by the strong law of large number, we only need to show $\dfrac{max_{1\leq k\leq n} |X_k|}{n} \to 0$ almost surely.
Firstly, let's remark that $\dfrac{X_n}{n} = \dfrac{S_n - S_{n-1}}{n} = \dfrac{S_n}{n} -\dfrac{n-1}{n}\dfrac{S_{n-1}}{n-1} \to EX_1 - EX_1 = 0$ almost surely by SLLN.
Then for any $\epsilon >0$, almost surely there exists $N$ such that $\left|{X_n}\right| < n\epsilon,\forall n > N$, which gives
$$\dfrac{\max_{1\leq k\leq n} |X_k|}{n} \leq \dfrac{\max_{1\leq k\leq N} |X_k| + n\epsilon }{n} \to \epsilon$$
i.e. $\limsup \dfrac{\max_{1\leq k\leq n} |X_k|}{n} \leq \epsilon , \forall \epsilon >0$, so $$\lim \dfrac{\max_{1\leq k\leq n} |X_k|}{n} = 0$$