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

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.

$$\frac{S_n}{\sqrt{2}\ln{n}}=\frac{\sqrt{2n\ln{\ln{n}}}}{\sqrt{2}\ln{n}}\frac{S_n}{\sqrt{2n\ln{\ln{n}}}}$$
$$\limsup_{n\to\infty}\frac{S_n}{\sqrt{2n\ln{\ln{n}}}}=1 \quad\text{a.s.,}$$
while the first term in the product converges to $+\infty$.