TITLE: Showing $\limsup \frac{S_n}{\sqrt{n}} = \infty$
QUESTION [7 upvotes]: Let $X_1, X_2, \dots$ be i.i.d random variables with $\mathbb{E}(X_i) = 0$ and $0 < Var(X_i) < \infty$. Use the Central Limit Theorem and Kolmogorov’s 0-1 Law to conclude that $\limsup \frac{Sn}{\sqrt{n}} = \infty$ a.s..
Kolmogorov's 0-1 Law: If $X_1, X_2, \dots$ are independent and $A \in \mathcal{T}$, then $\mathbb{P}(A) = 0$ or $\mathbb{P}(A) = 1$.
Let $A_x = \{{\limsup \frac{S_n}{\sqrt{n}}} > x\}$. Then $A_x \in \mathcal{T}$, so I just need to show that $\mathbb{P}(A_x) > 0$, because 
$\mathbb{P}( \limsup \frac{S_n}{\sqrt{n}} > x \ \text{i.o.}) \geq \mathbb{P}({\limsup \frac{S_n}{\sqrt{n}} > x}) = 1$.
I'm having some trouble to show that $\mathbb{P}(A_x) > 0$

REPLY [6 votes]: Use Fatou's lemma and the central limit theorem to get,
$$
\mathbb{P}\left(\limsup\frac{S_n}{\sqrt{n}}>x\right)\geq \limsup \mathbb{P}\left(\frac{S_n}{\sqrt{n}}>x\right)=\mathbb{P}(\mathscr{N}(0,var(X))>x)>0,
$$