Let $X_1, X_2, \dots$ be i.i.d random variables with $\mathbb{E}(X_i) = 0$ and $0 < \operatorname{Var}(X_i) < \infty$, then by the Central Limit Theorem and Kolmogorov’s 0-1 Law we can conclude that $$\limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt{n}} = \infty, \ a.s.$$
Now my question is that can we establish that $$\liminf_{n\rightarrow\infty} \frac{\vert S_n\vert}{\sqrt{n}} = 0, \ a.s.$$
It seems that the approach to the problem above fails here. I also tried to use the law of iterated logarithm to solve it, but only ended up with failure. Any help will be greatly appreciated.
Yes, intuitivelly because the modulus of Gaussian $\mathcal N(0,1)$ is concentrated on the whole $[0,\infty)$.
Precisely; assume contrary that $\liminf \frac{|S_n|}{\sqrt{n}} \neq 0$. By Kołmogorov $0-1$ Law we can ensure that for some $\varepsilon > 0$, $\liminf \frac{|S_n|}{\sqrt{n}} > \varepsilon$ almost surely. Then, by Fatou lemma, $$ 1=\mathbb P(\liminf \frac{|S_n|}{\sqrt{n}} > \varepsilon) = \mathbb E[ 1_{\{\liminf \frac{|S_n|}{\sqrt{n}}> \varepsilon\}}] \le \mathbb E[\liminf_n 1_{\{\frac{|S_n|}{\sqrt{n}} > \varepsilon\}}] $$ $$ \le \liminf_n \mathbb E[1_{\{\frac{|S_n|}{\sqrt{n}} > \varepsilon\}}] = \mathbb P(|\mathcal N(0,Var(X_1))| > \varepsilon) = 1 - 2\Phi(\frac{\varepsilon}{\sqrt{Var(X_1)}}) < 1.$$