# Almost sure convergence of a sum of independent r.v

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?

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 $$P( X_n=n^{2}-1 \ i.o.)=0$$.