Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of $L^2$ random variables in a probability space $(\Omega, \mathcal{A}, P)$. Define $\overline{X_n}:=\frac{1}{n}\sum_{i=1}^n X_i$. Show that (2) follows from (1):
(1) $\mathrm{Cov}(X_i,X_j)\leq c_{|i-j|}$ for a sequence $c_n$ with limit $0$.
(2) $\overline{X_n} $ converge in probability to the expected value of it (weak law of large numbers).
Without loss of generality I take $E(X_j)=0 \ \forall j$.
Be $ \epsilon>0$. It exists an $N: \forall n \ge N: c_n<\epsilon$. A convergent sequence is bounded: $\exists M: c_n< M < \infty \forall n \in \mathbb{N}$.
\begin{align} \mathrm{Var}(\overline{X_n}) &= \frac{1}{n^2} *( \sum_{i=1}^n \sum_{j=1}^n Cov(X_j,X_k)) \\ &= \frac{1}{n^2} *( \sum_{j,k: |j-k|\le N} Cov(X_j,X_k) + \sum_{j,k:|j-k|>N} Cov(X_j,X_k)) \\ &\le \frac{1}{n^2} * ( \sum_{j,k: |j-k|\le N}M + \sum_{j,k:|j-k|>N} *\epsilon) \\ &\le \frac{1}{n^2} (M*n*(1+2N)+ \epsilon* n^2) \\ &= \frac{M(1+2N)}{n}+ \epsilon \\ &\rightarrow \epsilon \quad (n\rightarrow \infty) \end{align}
Because I count for the entries with $|j-k|\le N $ the number $$n+2*(n-1)+2*(n-2)+..+2*(n-N)\\=n+2nN-2*(1+2+..+N)= n+2nN-2(\frac{(N+1)N}{2}) \le n*(1+2N)$$
I get $Var(\overline{X_n}) \rightarrow 0 \quad ( n \rightarrow \infty)$.
Be $\epsilon_1, \delta_1 >0$, so $ \exists N: Var(\overline{X_n}) < \epsilon_1^2 \delta_1 $ for $n \ge N $.
$$P(|\overline{X_n}|>\epsilon_1) \le \frac{Var(\overline{X_n})}{\epsilon_1^2} < \frac{\delta_1}{\epsilon_1^2} *\epsilon_1^2= \delta_1$$
I think i do some mistakes because the covariance don't must be positive so also the sequence? I think only if all is positive the proof is right.
Your approach works. The covariances may not be positive but it is not a problem when we write $$ \operatorname{Var}\left(\overline{X_n}\right)=\frac 1{n^2}\sum_{i,j=1}^n \operatorname{Cov}\left(X_i,X_j\right).$$ The sum is non-negative but the terms are not necessarily all non-negative .