$$\limsup_{n \to \infty} \frac{S_n}{\sqrt{n}}=+\infty, \quad\liminf_{n \to \infty} \frac{S_n}{\sqrt{n}}=-\infty \quad P\text{-a.s.}$$
Here $S_n$ is a random walk on $\mathbb{Z}$. I managed to show that the $\limsup_{n \to \infty} S_n \to +\infty$ $P\text{-a.s.}$ But, for this I have no idea how to deal with. Could anyone tell me how to show this fact?
Write $S_n=X_1+...+X_n$, where the $X_i$'s are your i.i.d. jumps with $\mathbb E[X_i]=0, \mathrm{Var}(X_i)=\sigma^2$ for $i=1,...,n$. Then the Central Limit Theorem tells you that $$\frac{S_n}{\sqrt {\sigma^2 n}} \to Z \sim \mathcal N(0,1)$$ from which you know that $Z$ is unbounded (as the tails of the normal distribution extend to infinity).