# Proof of Weak Law of Large Numbers with non-zero covariance

Let $$X_{1}, \cdots, X_{n}$$ be a sequence of dependent random variables having the same finite mean $$\mu=\mathrm{E}\left(X_{1}\right),$$ the same finite variance $$\sigma^{2}=\operatorname{Var}\left(X_{1}\right),$$ and $$\left|\operatorname{Cov}\left(X_{i}, X_{j}\right)\right| \leq \frac{1}{(i-j)^{4}}$$ for $$i \neq j .$$ Show that $$\bar{X}$$ converges in probability to $$\mu .$$

I am stuck on the upper bound of the sum of the covariances: $$\sum_{i=1}^{n} \sum_{j \neq i,j=1}^{n}|{\rm Cov}(X_i,X_j)|\le2 \cdot \sum_{k=1}^{n-1}(\frac{n-k}{k^4})$$

How to show that $$n^2$$ is a proper upper bound of the sum?

Since $$\left\lvert\operatorname{Cov}(X_i,X_j)\right\rvert\leq (i-j)^{-4}\leq 1$$, summing over all $$n(n-1)$$ ordered pairs $$(i,j)$$ with $$i\neq j$$ gives the easy upper bound $$n(n-1).
• But to make the probability tend to $0$, I need $n^2$ to be not asymptotically tight (the summation should not have a $n^2$ term). Feb 17, 2021 at 15:15
• In which case, $\sum_{k=1}^{n-1} (n-k)/k^4\leq (n-1)\zeta(4)=O(n)$. Feb 17, 2021 at 15:20