# Almost sure convergence of $|X_n|/n$ for a sequence of i.i.d r.v.'s

Let $$\{X_n: n \ge 1\}$$ be a sequence of i.i.d random variables with $$\mathbb{E}[|X_1|] < \infty$$, and $$\mathbb{E}[X_1] \neq 0$$. Show that $$\frac{|X_n|}{n} \to 0 \quad \text{almost surely.}$$ Use this result to show $$\frac{\max_{1\le k \le n}|X_k|}{S_n} \to 0 \quad \text{almost surely,}$$ where $$S_n := \sum_{j=1}^n X_j$$.

To show the first part, I tried to prove the equivalent result that $$\forall \epsilon>0 \quad \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon \quad \text{i.o.}\right) = 0,$$ which seemingly suggests the use of BC (1 or 2) lemma. The naive use of BC1 leads to nowhere since $$\sum_{n=1}^{\infty} \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\le \sum_{n=1}^{\infty}\frac{\mathbb{E}[|X_n|]}{n\epsilon} = \frac{\mathbb{E}[|X_1|]}{\epsilon} \sum_{n=1}^{\infty} \frac{1}{n} = \infty,$$ (which needs $$<\infty$$ to work). I also tried to use BC2: $$\sum_{n=1}^{\infty} \mathbb{P}\left(\frac{|X_n|}{n} \le \epsilon\right) = \sum_{n=1}^{\infty} \left(1- \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\right)$$ $$= \lim_{M \to \infty} M - \sum_{n=1}^{M} \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\ge \lim_{M \to \infty} M - \frac{\mathbb{E}[|X_1|]}{\epsilon} \sum_{n=1}^{M} \frac{1}{n} = \infty,$$ which results in $$\mathbb{P}\left(\frac{|X_n|}{n} \le \epsilon \quad \text{i.o.}\right) = 1,$$ again not helpful.

How can I crack the first part and also any idea for the second part is appreciated.

It is standard result that if $$Y$$ is a non-negative random variable then $$\sum_n P\{Y>n\} <\infty$$ iff $$EY<\infty$$. Taking $$Y=\frac {|X_1|} {\epsilon}$$ we get $$\sum P\{|X_n| >n\epsilon \}=\sum P\{|X_1| >n\epsilon \}<\infty$$. Apply Borel Cantelli Lemma and let $$\epsilon \to 0$$ through the sequence $$1,\frac 1 2,\frac 1 3,\cdots$$ to see that $$\frac {|X_n|} n \to 0$$ almost surely. If $$a_n \geq 0$$ and $$\frac {a_n} n \to 0$$ then $$\frac {\max \{a_1,a_2,\cdots,a_n\}} n \to 0$$. Now write $$\frac {\max \{|X_1|,|X_2|,\cdots, |X_n|\}} {S_n}$$ as $$\frac {\max \{|X_1|,|X_2|,\cdots, |X_n|\}} n \times \frac n {S_n}$$ and apply SLLN.