Show that $ {S_n\over 2^.5(logn)} $ converges to 0 almost surely.

Question

Suppose ${X_n, n ≥ 1}$ are independent identical random variables with $E(X_n) =0 $ and $E(X^2_n) =1 $ Show that $$ {S_n\over 2^.5(logn)} $$ converges to 0 almost surely. where $ S_n =\sum_{i=1}^n X_i $. I haven't got good idea how to solve it. I used SLLN. This doesn't work.

Answer 1

This is not true because

$$ \frac{S_n}{\sqrt{2}\ln{n}}=\frac{\sqrt{2n\ln{\ln{n}}}}{\sqrt{2}\ln{n}}\frac{S_n}{\sqrt{2n\ln{\ln{n}}}} $$

and by the LIL

$$ \limsup_{n\to\infty}\frac{S_n}{\sqrt{2n\ln{\ln{n}}}}=1 \quad\text{a.s.,} $$

while the first term in the product converges to $+\infty$.