Sum of squares of i.i.d. finite mean variables, divided by $n^2$, approaches $0$ a.s. 
Given $X_1,X_2,\dots$ are i.i.d, $X_n\ge0$, and $EX_1<\infty$, how can you show that
  $$
\frac{X_1^2+\dots+X_n^2}{n^2}\to 0\qquad \text{a.s.}?
$$

This is from an old probability qualifying exam. 
Some thoughts: You can show that $P(X_n\ge n\epsilon \text{ i.o})=0$, so that letting $Y_n=X_n1_{\{X_n\le n\epsilon\}}$, and $T_n=\sum_1^n Y_k^2$, it suffices to show $\frac{T_n}{n^2}\to 0$. Using Borel-Cantelli, to prove that it suffices to show $\sum_1^\infty P(T_n>\epsilon n^2)<\infty$ for all $\epsilon>0$. It's hard to get a good bound on $P(T_n>\epsilon n^2)$, though, since $ET_n$ grows as $n^3$. I imagine some kind of Holder/Jensen's inequality trick is necessary, though I cannot see it. Does anyone have any ideas?
My question is equivalent the $p=1/2$ case of the $(\Longleftarrow)$ part of this unsolved question. This may make my question a duplicate, but I figured that mine being a special case might mean it was easier.
 A: This question can be solved by standard machinery, the Kronecker's Lemma and Kolmogorov's three-series theorem. 
By using Kronecker's lemma, we only need to show that $$\sum \frac{X_n^2}{n^2} <\infty.$$ 
Now by the three series theorem, we only need to show that $Y_i =\frac{X_i^2}{n^2}1_{\{|X_i|< n\}}$ satisfies the  three series conditions. 
(1) $\sum_{n=1}^{\infty} P(|X_i|>n)=\sum_{n=1}^{\infty} P(|X_1|>n)$ is finite by noting that $X_i$ are iid and the first moment is finite and the equivalence between $E(|X_1|)<\infty \Leftrightarrow  \sum P(X_1>n)$. 
(2)$\sum E(Y_i)=\sum E(\frac{X_i^2}{n^2}1_{\{|X_i|< n\}})=\sum E(\frac{X_1^2}{n^2}1_{\{|X_1|< n\}})= \sum_{n=1}^{\infty}\sum _{k=1}^{n} E(\frac{X_1^2}{n^2}1_{\{k-1\leq|X_1|< k\}})=\sum_{k=1}^{\infty} EX_1^2 1_{\{k-1\leq |X_1|\leq k\}}\sum_{n=k}^{\infty} n^{-2} \leq \sum_{k=1}^{\infty}\frac{2}{k} EX_1^21_{\{k-1\leq |X_1|\leq k\}} \leq 2\sum_{k=1}^{\infty} EX_1 1_{\{k-1\leq |X_1|\leq k\}}=2EX_1<\infty$. 
I use $\sum_{n=k}^{\infty} n^{-2}\leq 2/k $ above by consider the integral $\int_{k}^{\infty} t^{-2} dt$ and  the assumption $X_n\geq 0.$ 
(3) $\sum \text{Var}(Y_i) \leq \sum E(\frac{X_1^4}{n^4} 1_{\{X_i<n\}})$. 
Use the same trick $\int_{k}^{\infty} t^{-4}dt\leq C/k^3$, I think you can show the sum is finite as well. 
