Prove $\liminf(M_n = -1) \subseteq ((\lim \frac{S_n}{n}) = -1)$ 
Let $M_1, M_2, \dots$ be iid RVs s.t. $P(M_n = n^2 - 1) = \frac{1}{n^2} = 1 - P(M_n = - 1)$. Define $S_n = \sum_{i = 1}^{n} M_i$. Prove that $P\{(\lim \frac{S_n}{n}) = -1\} = 1$.

$$\sum_n P(M_n = n^2 - 1) = \sum_n \frac{1}{n^2} < \infty$$
By BCL1, $$\to P\{\limsup(M_n = n^2 - 1)) = 0$$
$$\to P\{\liminf(M_n = -1)\} = 1$$
It seems intuitive, but how do we show rigorously that $$P\{(\lim\frac{S_n}{n}) = -1\} = 1?$$?
 A: Let's denote $\liminf_n[M_n = -1]$ by $A$, by first Borel-Cantelli lemma, $P(A) = 1$. Therefore for each $\omega \in A$, there exists $N \in \mathbb{N}$, such that $M_n(\omega) = -1$ for all $n > N$. It then follows that
\begin{align}
& \left|\frac{S_n(\omega)}{n} - (-1)\right| \\
= & \left|\frac{M_1(\omega) + \cdots + M_N(\omega)  - N \times (-1)}{n} \right|\\
\leq & \frac{1^2 - 1 + 2^2 - 1 + \cdots + N^2 - 1 + N}{n} \\
= & \frac{\frac{1}{6}N(N + 1)(2N + 1)}{n} \\
\to & 0
\end{align}
as $n \to \infty$. This shows that for each $\omega \in A$, $\lim\limits_{n \to \infty}\frac{S_n(\omega)}{n} = -1$. In other words, $\frac{S_n}{n} \to -1$ with probability $1$.
A: Let $\omega \in \Omega$. If $\omega \in \{\liminf(M_n = -1)\}$,
then $\exists m \ge 1$ s.t. $-1 = M_m(\omega) = M_{m+1}(\omega) = ...$
Observe that $$(\frac{S_{m+k}}{m+k})_{k \ge 0}$$ is a decreasing sequence bounded below by -1.
Hence,
$$\lim_{k \to \infty} \frac{S_{m+k}}{m+k} = - 1$$
$$\lim_{k \to \infty} \frac{S_{m+k}}{m+k} = - 1$$
$$\lim_{k \to \infty} \frac{S_{k}}{k} = - 1$$
$$\lim_{n \to \infty} \frac{S_{n}}{n} = - 1$$
Is that right?
