Suppose $(X_i)_{i\in\mathbb{Z}}$ is a strongly mixing, strictly stationary sequence of random variables. If we assume $\mathbb{E}[|X_i|^{2+\epsilon}]<\infty$ for some $\epsilon>0$ and if we have for the mixing coefficients that they satisfy for $k$ large enough $\alpha(k) := \alpha(\sigma((X_i)_{i\leq 0}), \sigma((X_i)_{i\geq k})) \leq Ck^{-\gamma}$ for some $C>0$ and $\gamma\geq \frac{(1+\epsilon)(2+\epsilon)}{\epsilon}.$ Then Davydov's inequality implies that
$$|Cov(X_0, X_k)|\leq 8||X_1||_{2+\epsilon}^2 [\alpha(k)]^{\epsilon/(2+\epsilon)}\leq 8C||X_1||_{2+\epsilon}^2 k^{-(1+\epsilon)}.$$ This then implies that the autocavariances are absolutely summable: $$\sum_{j=0}^\infty |Cov(X_0, X_j)|<\infty.$$
It seems to me that there must be some counterexample possible if we only have $\mathbb{E}[X_i^2]<\infty$, rather than $\mathbb{E}[|X_i|^{2+\epsilon}]<\infty$, even with polynomially decaying mixing coefficients $\alpha(k)\leq Ck^{-(3+2\sqrt{2})}$ for some $C>0$ (for $k$ large enough). The $3+2\sqrt{2}$ is the minimal value of $(\epsilon+1)(\epsilon+2)/\epsilon$ for $\epsilon>0$. However, I have difficulties coming up with a counterexample myself. What would be an example of such a (strongly mixing, strictly stationary) sequence that does not have absolutely summable autocovariances?