I have a problem from probability theory that I have encountered and I am not getting anywhere in terms of ideas that may help me solve it.
Suppose $\{X_{n, i}\}_{1 \leq i \leq n, n \geq 1}$ are mutually independent with a common marginal distribution, given by $P\left(X_{1,1}=n\right)=R n^{-3}$ for every $n \geq 1$, where $R$ is the normalizing constant. Let $S_{n}=\sum_{i=1}^{n} X_{n, i}$. Show that $E\left|X_{1,1}\right|<\infty$, but
$\limsup \frac{S_{n}}{n}=\infty$ almost surely
The expectation is found with simple algebra, however, when I am trying to determine that $\limsup \frac{S_{n}}{n}=\infty$ I go blank.
I originally thought of considering as $A_n$ the set where $\frac{S_n}{n}$ is greater than $n$, that is $P(A_n)=P(\frac{S_n}{n}>n)$.
Then I was thinking that if we could consider the infinite sum $\sum_{n=1}^{\infty}P(A_n)$ and show that this diverges, we can apply the Borel-Cantelli lemma. I don't really know how to obtain a convenient lower bound for $P(A_n)$ for this problem since what I have used before is Markov's inequality, but when using it here we would be with an upper bound.
I am only looking for a hint as to what may be useful to approach this question.
Note that for some $c>0$ and all large $k$ $$ \mathbb{P}(X\ge k)\ge \frac{c}{k^2} $$ (since $\sum_{n=k}^\infty n^{-3}\sim k^{-2}$). Therefore, $$ \mathbb{P}(X_{n,i}\ge n\sqrt{\ln n})\ge \frac{c}{n^2\ln n} $$ and if $A_n=\left\{\frac{S_n}n\ge \sqrt{\ln n}\right\}$, then $$ \mathbb{P}(A_n)\ge\mathbb{P}(X_{n,i}\ge n\sqrt{\ln n}\text{ for some }i=1,2,\dots,n)\ge 1-\left(1-\frac{c}{n^2\ln n}\right)^n = \frac{c+o(1)}{n\ln n} $$ Now since the events $A_n$ are independent and $\sum\mathbb{P}(A_n)=\infty$, by the second Borel-Cantelli lemma the events $A_n$ occur infinitely often a.s. This implies that $$ \frac{S_n}{n}\ge \sqrt{\ln n} $$ infinitely often a.s., implying the required result.