Almost sure convergence of a sum of independent r.v

Question

Let $S_n:=\sum_{i=1}^nX_i$ where $X_1,X_2,...$ are indepentent r.v.'s such that:

$P(X_n=n^2-1)=\frac{1}{n^2}$ and $P(X_n=-1)=1-\frac{1}{n^2}$

Show that $\frac{S_n}{n}\rightarrow-1$ almost sure.

It is easy to see that you can't apply strong law of large numbers (SLLN) because: $\forall i\neq j: E(X_i)=E(X_j)$. Which is why it is not converging a.s. to $E(S_n)=0$.

Now i thought about applying the Borel-Cantelli-Lemma on ...

$\{\lim_{n\rightarrow\infty}S_n=-1\}=\bigcap_{k\geq1}\bigcup_{m\geq1}\bigcap_{n\geq m}\{\omega\mid |\frac{1}{n}\sum_{i=1}^nX_i+1|\leq\frac{1}{k}\}$

... but did not succeed. So how do I prove almost sure convergence here?

Thanks in advance.

Answer 1

If you have a sequence $(a_n)$ of real numbers such that $a_n=-1$ for all but finitely many values of $n$ then it is easy to check that $\frac 1 n \sum\limits_{k=1}^{n} a_k \to -1$. So we only have to show that $X_n=-1$ for all $n$ sufficiently large, with probability $1$. But this follows from Borel - Cantelli Lemma since $\sum P(X_N=n^{2}-1)=\sum \frac 1{n^{2}}<\infty$ so $P( X_n=n^{2}-1 \ i.o.)=0$.