Series of autocovariances is nonnegative Define the autocovariance function $\gamma_X$ of a stationary time series $X$ by $\gamma_X(k) = L_\gamma(k) \vert k \vert^{-\theta}$, $k \in \mathbb{Z}$, where $L_\gamma$ is slowly varying at infinity and $\theta \in (0, 2)$.
Is it true that
$$
\sum_{k \in \mathbb{Z}} \gamma_X(k) \geq 0
$$
I suspect that this is true and can be shown by using the semi-positive definiteness of any covariance function but I cannot get the LHS of the inequality into the bilinear form such that I can use the semi-positive definiteness.
EDIT: Note that $\gamma(0)$ is equal to the variance $\sigma$ of $X(1)$ and therefore $L(0)$ is known. Therefore, the question is whether the remaining covariances can sum up to something smaller than $-\sigma$. I suspect that the semipositive definiteness of $\gamma$ implies that they can't.

As you probably know, the sum does not even converge unless $\theta>1$.
But it turns out that $\sum_n \gamma(n)\ge 0$ for any positive definite $\gamma$, assuming that $\sum_n e^{2\pi i nx}\gamma(n)$ is a continuous function for $x\in [0,1]$ (I believe this is the case for your $\gamma$s, assuming that $\theta>1$).
Why? According to Bochner's Representation Theorem,the Fourier transform $\hat{\gamma}(x)=\sum_n e^{2\pi i nx}\gamma(n)$ is a nonnegative function of $x$ (here is where we use the nonnegative-definiteness). In particular, $\sum_n \gamma(n)=\hat{\gamma}(0)$.
