Cfrgtkky

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 $$frac{max_{1le k le n}|X_k|}{S_n} to 0 quad text{almost surely,}$$ 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 $$sum_{n=1}^{infty} mathbb{P}left(frac{|X_n|}{n} > epsilonright)le sum_{n=1}^{infty}frac{mathbb{E}[|X_n|]}{nepsilon} = frac{mathbb{E}[|X_1|]}{epsilon} sum_{n=1}^{infty} frac{1}{n} = infty,$$
(which needs $<infty$ to work). I also tried to use BC2:
$$sum_{n=1}^{infty} mathbb{P}left(frac{|X_n|}{n} le epsilonright) = sum_{n=1}^{infty} left(1- mathbb{P}left(frac{|X_n|}{n} > epsilonright)right)$$
$$= lim_{M to infty} M - sum_{n=1}^{M} mathbb{P}left(frac{|X_n|}{n} > epsilonright)ge lim_{M to infty} M - frac{mathbb{E}[|X_1|]}{epsilon} sum_{n=1}^{M} frac{1}{n} = infty, $$
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 $sum P{|X_n| >nepsilon }=sum P{|X_1| >nepsilon }<infty$. 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.