Suppose let $\displaystyle (X_n)_{n\ge 1}$ be i.i.d. with N(1,3) random variables. How can I show that:

$\displaystyle \displaystyle\lim_{n\to\infty}\frac{X_1+...+X_n}{X _1^2+...+X_n^2}=\frac{1}{4} $ (a.s)

From my understanding: $\displaystyle \displaystyle\lim_{n\to\infty}\frac{S_n}{n}=\displ aystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\s um_{j=1}^{n}X_j=\mu$

Then the nominator will go to 1.. How about the denominator? Why is its expected value 4?