Original problem:
Let $X_1, X_2, ...$ be random variables in $(\Omega, \mathscr{F}, \mathbb{P})$
where $X_1 = 0$ and for $n > 1$
$$P(X_n=-n^2)=\frac{1}{n^2}$$
$$P(X_n=-n^3)=\frac{1}{n^3}$$
$$P(X_n=2)=1-\dfrac{1}{n^2}-\dfrac{1}{n^3}$$
Let $S_n=\sum_{i=1}^{n} X_i$. Prove that $P([\lim S_n] = \infty) = 1$
I used BCL1 to deduce that
$$P(\liminf [X_n = 2]) = 1$$
Let $\omega \in \Omega$. If $\omega \in \liminf [X_n = 2]$, then $\exists m \ge 1$ s.t.
$$2 = X_m(\omega) = X_{m+1}(\omega) = ...$$
$$\to S_{m+k}(\omega) = S_m(\omega) + 2k \ \forall k \ge 0$$
$$\to \lim _{k \to \infty} S_{m+k}(\omega) = \infty$$
$$\to \lim _{k \to \infty} S_{k}(\omega) = \infty$$
$$\to \lim _{n \to \infty} S_{n}(\omega) = \infty$$
The result follows by monotonicity of probability. QED