Show that a sequence of Random Variables $(X_i)_{i \in \mathbb{N}}$ obeys the weak law of large numbers given certain conditions on $\rho(X_i,X_j)$ Say I have a sequence of Random (real-valued) Variables $(X_i)_{i \in \mathbb{N}}$ such that $E[X_i] = \mu, Var(X_i)=\sigma^2$. Furthermore say there exists one $k \in \mathbb{N}$ such that  $\rho(X_i,X_j) = 0$ for $|i-j|>k$. a.k.a. the random variables are uncorrelated as soon as they are far enough apart in the sequence ($k$ apart, to be precise.)
Now I want to show, that the sequence $(X_i)_{i \in \mathbb{N}}$  obeys the weak law of large numbers, a.k.a. $$\lim_{n \to \infty} P \left( \left|\frac{1}{n}\sum_{k=1}^{n}(X_k-E[X_k]) \right| \geq \epsilon \right)= 0$$
My guess is that I have to tweak the standard proof somehow. There are proofs available here on stackexchange for similar hypotheses, but none quite like the one I'm searching for. 
 A: Hints:


*

*Show (or recall) that $$\mathbb{E} \left( \left[ \sum_{i=1}^n (X_i-\mu) \right]^2 \right) = \sum_{i=1}^n \sum_{j=1}^n \varrho(X_i,X_j).$$ 

*For each fixed $i$ there are at most $2k$ choices for $j$ such that $\varrho(X_i,X_j) \neq 0$. Moreover, by the Cauchy-Schwarz inequality, $$|\varrho(X_i,X_j)| \leq \sigma^2$$ for all $i,j \geq 1$.

*Conclude that $$\mathbb{E} \left( \left[ \sum_{i=1}^n (X_i-\mu) \right]^2 \right) \leq 2kn \sigma^2.$$

*Apply the Markov inequality to prove that $$\mathbb{P} \left( \left| \frac{1}{n} \sum_{i=1}^n (X_i-\mu) \right|> \epsilon \right) \leq \frac{1}{n^2 \epsilon^2} \mathbb{E} \left( \left[ \sum_{i=1}^n (X_i-\mu) \right]^2 \right).$$

*Conclude.

A: $\newcommand{\P}{\mathbb{P}}
\newcommand{\E}{\mathbb{E}}
\newcommand{\Var}{\text{Var}}
\newcommand{\Cov}{\text{Cov}}$
So, you want to tweak the standard proof somehow? Okay. Recall the Chebyshov's inequality first:
\begin{align}
\P(|X|\geq a)\leq \frac{\E[X^2]}{a^2}
\end{align}
For $a>0$. 
Also recall that:
\begin{align}
\Var\left( \sum_{i=1}^nX_i\right) = \sum_{i=1}^n\sum_{j=1}^n \Cov(X_i,X_j)
\end{align}
We have every tool that we need right now.
\begin{align}
\P\left(\bigg|\frac{1}{n}\sum_{i=1}^n(X_i-\mu)\bigg|\geq \epsilon\right)\leq \frac{\E\bigg[\frac{1}{n^2}\left(\sum_{i=1}^n(X_i-\mu)\right)^2\bigg]}{\epsilon^2}
\end{align}
The RHS can be written as:
\begin{align}
\frac{\E\bigg[\frac{1}{n^2}\left(\sum_{i=1}^n(X_i-\mu)\right)^2\bigg]}{\epsilon^2} = \frac{1}{\epsilon^2 n^2}\Var\left( \sum_{i=1}^nX_i\right)  = \sum_{i=1}^n\sum_{j=1}^n \Cov(X_i,X_j)
\end{align}
Now recall from probability theory that $\rho(X,Y)=\frac{\Cov(X,Y)}{\sqrt[]{\Var(X)\Var(Y)}}$. Hence for  random variables with nonzero variance we have $\rho=0$ iff $\Cov=0$. So:
\begin{align}
\sum_{i=1}^n\sum_{j=1}^n \Cov(X_i,X_j) = \sum_{i=1}^n\sum_{\min\{-k+i,1\}}^{k+i} \Cov(X_i,X_j)
\end{align} 
We also have $\Cov(X,Y)\leq \sqrt[]{\Var(X)\Var(Y)}$ by Cauchy-Schwarz. So:
\begin{align}
\sum_{i=1}^n\sum_{\min\{-k+i,1\}}^{k+i} \Cov(X_i,X_j) \leq 2kn\sigma^2
\end{align}
Putting everything together:
\begin{align}
\P\left(\bigg|\frac{1}{n}\sum_{i=1}^n(X_i-\mu)\bigg|\geq \epsilon\right)\leq  \frac{2k\sigma^2}{n\epsilon^2}
\end{align}
Taking the limit gives the answer.
