Sorry I posted this question on a new account I created because I was not able to gain access to my main acct. I was able to get back on to my main acct.
I have been self studying probability using some Stanford lecture notes and am trying to see if my logic is correct for the following exercise:
$X_{i}$ are mutually independent with $P(X_n=n^2-1)=1-P(X_n=-1)=\frac{1}{n^2}$ then $\mathbb{E}[X_n]=0$ and
$n^{-1}\sum_{i=1}^{n}X_{i}$ converges almost surely to -1. I figured since $P(X_n= -1)\rightarrow 1$ then $P(X_n= -1\hspace{3mm} a.a)=1$ More formally:
$$P(\cap_{n=1}\cup_{k\ge n}|X_n+1|>\epsilon)\le \sum_{k\ge n}P(|X_n+1|>\epsilon)$$
then taking the limit on both sides shows $$P(\cap_{n=1}\cup_{k\ge n}|X_n+1|>\epsilon)=0$$ since $P(|X_n+1|>\epsilon)\rightarrow 0$ then once can conclude $X_n\rightarrow -1$ almost surely then the convergence of the sum follows. Does any of this make sense?
$P(|X_n+1| >\epsilon \to 0$ for every $\epsilon >0$ only tells you that $X_n \to -1$ in probability. How does this give a.s convergence?
Here is a simple argument using Borel-Cantelli Lemma:
$\sum P(X_n=n^{2}-1) <\infty$ so there is probability $0$ that $X_n=n^{2}-1$ for infinitely many $n$. This implies that $X_n=-1$ for all $n$ sufficiently large, with probability $1$. This in turn implies that $\frac 1 n \sum\limits_{k=1}^{n} X_k \to -1$ a.s..