Imagine we have given a stationary process $\{X_t\}_{t \in \mathbb Z}$;
Now I want to show that
$$\operatorname{Var}(X_1+\cdots+X_T) \to \infty,$$ for $T \to \infty$.
We have that the process if $\phi$-mixing with $\phi(n) \leq n^{-(1+\varepsilon)}$ for some $\varepsilon >0$.
For those who don't know the meaning of $\phi$-mixing: You can see it here: https://www.encyclopediaofmath.org/images/a/a7/Strong_mixing_conditions.pdf, Page 2;
Now I don't have a clear plan; I could say:
$$\operatorname{Var}(X_1+\cdots+X_T)=\sum_{i=1}^T \sum_{j=1}^T \operatorname{Cov}(X_i,X_j)$$
The variance of each element is the same and strictly greater than zero since the process is not constant; Now I somehow should use the mixing coefficient to show that the variance of the sum really goes to infinity;
One idea is using $\rho(n)\leq 2\sqrt{\phi(n)}$, where $\rho$ is defined again as in:
https://www.encyclopediaofmath.org/images/a/a7/Strong_mixing_conditions.pdf, Page 2
(since $\operatorname{Cov}(X_1,X_j)\geq -\rho(\vert i-j\vert)\operatorname{Var}(X_1)$)
However then the decay-rate of the $\phi$-mixing coefficient is not enough; Do you have any other idea to show the statement?