Problem
Let $X_n$ be a sequence of mixing random variables. If there is $a>1$ such that the strong mixing coefficient satisfies $$\alpha(s)\leq Cs^{-a}\quad (1)$$ ($C>0$ is a generic positive constant) then we can use a probability inequality of the form
$$P(\lvert \sum_{i=1}^n X_i\rvert>\epsilon)\leq f(n,a),\quad (2)$$
where $f$ is a function. I want to show that $\sum_{n=1}^\infty P(\lvert \sum_{i=1}^n X_i\rvert>\epsilon)< +\infty$ . However, in my case, the decaying rate of $\alpha(s)$ is faster, i.e., $$\alpha(s)\leq Ct^{s}, t\in(0,1). \quad (3)$$ It is clear that $(3)$ implies $(1)$ for any $a>0$, and so $(2)$ is a possible way to try to bound the probability.
The problem is that for small values of $a>0$, $\sum_{n=1}^\infty f(n,a)=+\infty$, and for large values of $a$, $\sum_{n=1}^\infty f(n,a)\leq+\infty$. Does it mean that this proof strategy failed?
Discussion
I believe this is more a matter of logic. On one hand, under the geometric decay rate in $(3)$, I can show that if we pick $a$ large enough so that $(1)$ is true, then $\sum_{n=1}^\infty f(n,a)\leq+\infty$ as desired. On the other hand, I can show that $(3)$ implies that there is a small $a$ so that $(1)$ is true and $\sum_{n=1}^\infty f(n,a)=+\infty$ which does not prove what I want. So, if I want to show that $$\sum_{n=1}^\infty P(\lvert \sum_{i=1}^n X_i\rvert>\epsilon)< +\infty,$$ given $(3)$, is it sufficient to just pick $a$ large enough?
It is strange to conclude that the series of probabilities converges when things (dependence) decay slowly and and does not converge when the dependence decays faster...My intuition says that if I know that the desired convergence holds for any $a>3$, then under geometric mixing rate, I can also pick any $a>3$ (and ignore the smaller values of $a$).
Can you give me feedbacks about it?
*The inequality $(2)$ is the Fuk-Nagaev's inequality Theorem 6.2 of Rio.