Let $(X_n)_{n\geq1}$ be i.i.d. real-valued random variables such that $\mathbb{E}(X_1)=0$ and $\mathrm{Var}(X_1)=\mathbb{E}({X_1}^2)>0$, where the variance $\mathrm{Var}(X_1)$ can be $+\infty$. Let $(S_n)_{n\geq1}$ be a random walk of $(X_n)$, i.e. $$S_n:=\sum_{k=1}^{n}X_k.$$ Note that the expectation and variance of $S_n$ is $$\mathbb{E}(S_n)=0,\qquad \mathrm{Var}(S_n)=\mathbb{E}({S_n}^2)=n\mathrm{Var}(X_1)=n\mathbb{E}({X_1}^2)>0.$$ Then I want to prove the following: $$\limsup_{n\to\infty}S_n=+\infty\quad \mathrm{a.s.}\qquad\mathrm{and}\qquad\liminf_{n\to\infty}S_n=-\infty\quad \mathrm{a.s.}$$ In my attempt, I considerd $\limsup$ and $\liminf$ separately. First, we get $$\mathbb{P}\left(\limsup_{n\to\infty}S_n>\varepsilon\right)=0\quad \mathrm{or}\quad 1$$ for any $\varepsilon>0$ from the Kolmogorov's 0–1 law. If that value is $0$, then we have $$\mathbb{P}\left(\limsup_{n\to\infty}S_n\leq\varepsilon\right)=1$$ i.e. $$\limsup_{n\to\infty}S_n\leq\varepsilon\quad \mathrm{a.s.}$$ Thus $$\limsup_{n\to\infty}S_n\leq0\quad \mathrm{a.s.}\tag{1}$$ by $\varepsilon\to+0$. Also, from the Fatou's lemma, we obtain $$0=\limsup_{n\to\infty}\mathbb{E}(S_n)\leq\mathbb{E}\left(\limsup_{n\to\infty}S_n\right)\leq0.$$ Therefore $\displaystyle\limsup_{n\to\infty}S_n=0$ a.s. because of (1).
My proof is stopped here. I want to lead a contradiction by using $\mathrm{Var}(S_n)\to +\infty$, but I don't know the concrete idea. Would someone help me? Thank you.