Let $(X_t)$ be a stationary process and let $\gamma(\cdot)$ denote its autocovariation function.
Statement If $\gamma(0)>0$ and $\gamma(h)\to 0$ as $h\to\infty$, then the covariance matrix $\Gamma_n$ is non-singular for every $n$.
The proof is by contradiction. Assume that there is some $n$ such that $\Gamma_n$ is singular. Then it can be shown (I omit that here) that there is an $r\geq 1$, such that for each $n\ge r+1$, $$ \gamma(0)\leq\sum_{j=1}^r \lvert a_j^{(n)}\rvert\cdot\lvert\gamma(n-j)\rvert, $$
where, for each fixed $j$, $a_j^{(\cdot)}$ is a bounded function of $n$. Hence, setting $M:=\max_{1\le j\le r}\left\{M_j\right\}$ with $\lvert a_j^{(n)}\rvert\leq M_j$ for each $n\ge r+1$, we get $$ \gamma(0)\leq M \sum_{j=1}^r \lvert\gamma(n-j)\rvert.~~~~(*) $$
It is said that this shows that it is not possible that both $\gamma(0)>0$ and $\gamma(h)\to 0$ as $h\to\infty$ are fulfilled.
I am not sure if I did understand this last conlusion.
I think this is meant the following way. Since $(*)$ holds for each $n\ge r+1$, we can consider $n\to\infty$. If we let $n\to\infty$, then $n-j\to\infty$ for each summand. So if $\gamma(h)\to 0$ as $h\to\infty$, the right hand side converges to $0$ as $n\to\infty$ and we get $\gamma(0)=0$.
So both $\gamma(0)>0$ and $\gamma(h)\to 0, h\to\infty$ cannot be fulfilled simultaneuosly.
Is that the way it may be meant?