If $X_1,X_2,…$ is a sequence of independent random variables such that, \begin{cases} n^2-1 & \text{with probability } \frac{1}{n^2} \\ -1 & \text{with probability } 1-\frac{1}{n^2} \end{cases} How do I show that for every $n$ we have $E(X_n)=0$, but almost surely we have $\frac{X_1+X_2+\dots+X_n}{n}\longrightarrow-1$ as $n\rightarrow\infty$.
This is what I've managed to do.
First, let's calculate the expectation of $X_n$. \begin{align*} E(X_n) = (n^2-1)\frac{1}{n^2}+(-1)(1-\frac{1}{n^2}) &= 1-\frac{1}{n^2}-1+\frac{1}{n^2} \\ &= 0 \end{align*} Given the $P(X_n\in\{n^2-1,-1\})=1$ and $P(X_n=n^2-1)=\frac{1}{n^2}$ and $P(X_n=-1)=1-\frac{1}{n^2}$ \begin{align*} &\sum^\infty_{n=1}P(X_n=n^2-1) = \sum^\infty_{n=1}\frac{1}{n^2}\rightarrow\frac{\pi^2}{6} \\ &\sum^\infty_{n=1}P(X_n=-1) = \sum^\infty_{n=1}1-\frac{1}{n^2}\rightarrow\infty \end{align*} Now, $P(X_n=-1)\longrightarrow1$ as $n\rightarrow\infty$ by Borel Cantelli, since $P(\limsup(X_n=n^2-1))\rightarrow0$. Following this, \begin{align} \lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_k)=\lim_{n\rightarrow\infty}\big(\sum^{n^*}_{k=1}\frac{X_k}{n}+\sum^{n}_{k=n^*}\frac{-1}{n} \big) \longrightarrow -1 \end{align} where, $n^*$ is such that $X_n=-1$ for $n>n^*$ and $n^*>0$. \ \ As $n\rightarrow\infty, \sum^{n^*}_{k=1}\frac{X_k}{n}\rightarrow0$ and $\sum^{n}_{k=n^*}\frac{-1}{n}=\frac{n^*}{n}-1$. Therefore, \begin{align*} \lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_n)\longrightarrow -1 \end{align*}
