LLN question: convergence of sequence average This is a question relation to the law of large numbers. Let $(X_n:N\in\mathbb{N})$ is as sequence of independent random variables such that $\mathbb{E}(X_n) = \mu$ and $\mathbb{E}(X_n^4) \leq M$ for all n, for some constants $\mu \in \mathbb{R}$ and $M<\infty$. Now set $P_n = X_1X_2+X_2X_3+...+X_{n-1}X_n$. Then I wish to show that $P_n$/$n$ converges almost surely as $n\to\infty$, and I wish to find this limit.
Thanks you so very much!!
This is a homework question, and in class we have looked at the condition $\mathbb{E}(X_n^4) \leq M$ in the context of the law of large numbers, so I'm quite sure this should come in at some point.
My thoughts are the series $P_n$ / $n$ converges to $\mu^2$, but I am uncertain what I need to state to show this.
 A: Recall the following theorem:

Let $(X_n)_{n \in \mathbb{N}} \subseteq L^2(\mathbb{P})$ be a sequence of independent random variables such that $$\sum_{n=1}^{\infty} \frac{\mathbb{V}X_n}{n^2} < \infty$$ where $\mathbb{V}X_n$ denotes the variance of $X_n$. Then, $$\frac{1}{n} \sum_{i=1}^n (X_i-\mathbb{E}X_i) \to 0 \quad \text{a.s.}$$


Since the random variables are independent and $\mathbb{E}X_j = \mu$, we have
$$\mathbb{E}((X_j+X_{j-1})^2) = \mathbb{E}(X_j^2)+\mathbb{E}(X_{j+1}^2)+2\mu^2$$
hence,
$$\begin{align*}2 X_j X_{j+1} &= (X_j+X_{j+1})^2 - X_j^2-X_{j+1}^2 \\ &= \big[ (X_j+X_{j+1})^2 - \mathbb{E}((X_j+X_{j+1})^2) \big] - \big[ X_j^2-\mathbb{E}(X_j^2) \big] - \big[ X_{j+1}^2 - \mathbb{E}(X_{j+1}^2) \big] +2 \mu^2\end{align*}$$
Therefore, we can write
$$P_n = Q_n^1+Q_n^2+Q_n^3+(n-1)\mu^2 \tag{1} $$
where
$$\begin{align*}
Q_n^1 &:= \frac{1}{2} \sum_{j=1}^{\lfloor \frac{n}{2} \rfloor} \bigg[(X_{2j}+X_{2j+1})^2-\mathbb{E}((X_{2j}+X_{2j+1})^2) \bigg] \\
Q_n^2 &:=\frac{1}{2} \sum_{j=1}^{\lfloor \frac{n}{2} \rfloor} \bigg[(X_{2j+1}+X_{2j+2})^2-\mathbb{E}((X_{2j+1}+X_{2j+2})^2) \bigg] \\
Q_n^3 &:= \sum_{j=1}^n \bigg[ X_j^2- \mathbb{E}(X_j^2) \bigg]+ \frac{1}{2} (X_n^2-\mathbb{E}(X_n^2)) \end{align*}$$
By the previous theorem and the assumption $\mathbb{E}(X_n^4) \leq M$, we know that
$$\lim_{n \to \infty} \frac{1}{\lfloor \frac{n}{2} \rfloor} Q_n^1 = \lim_{n \to \infty} \frac{1}{\lfloor \frac{n}{2} \rfloor} Q_n^2 = \lim_{n \to \infty} \frac{1}{n} Q_n^3 = 0 \quad \text{a.s.} \tag{2}$$
Combining $(1)$ and $(2)$, we find
$$\frac{P_n}{n} \to \mu^2 \quad \text{a.s.}$$
A: $E(X_nX_{n+1})=E(X_n)E(X_{n+1})=\mu^2$ by the independence.
Then 
$E(P_n)=n\mu^2$
