I'm studying a Central Limit Theorem obtained by Politis and Romano (p.309, 1997) and there says: "by assumption (A.9) we may assume, without loss of generality, that $\alpha_n(k)\leq K/k^2$".
Assumption $(A.9)$ is $\sum_{k=0}^\infty (k+1)^2[\alpha_n(k)]^{\delta/(\delta+4)}\leq K,$ for all $n$, and for some $\delta>0$. It is good to know that the $\alpha$-mixing coefficient respects the bounds $0\leq \alpha_n(.)\leq1/4$.
Do you have any idea why we can pick any $\alpha_n(k)\leq K/k^2$, W.L.O.G.?
Thanks in advance!
UPDATE
Let's pick $\alpha_n(k)= K/k^2$. If the statement is correct, this choice should satisfy $\sum_{k=0}^\infty (k+1)^2[\alpha_n(k)]^{\delta/(\delta+4)}\leq K$, which implies that $\lim_{k\to\infty}(k+1)^2[\alpha_n(k)]^{\delta/(\delta+4)}=0$. But note that $(k+1)^2[K/k^2]^{\delta/(\delta+4)}$ diverges! The contraposition says that the series should not converge. This is an contradiction.
Am I misunderstanding what the author intended to say, or it is indeed wrong?
This was tricky. My interpretation is that he purposely allowed the mixing coefficient to decrease slow enough so that the Assumption (A.9) does not hold in some cases (i.e., the series is divergent). By doing so, he showed that the next result $$16r_n\alpha_n(l_n)\leq O(d_n^{-1/4})=o(1),$$ that is, this sequence converges to zero. If this result holds for $\alpha_n(k)=O(k^{-2})$, then it certainly holds for faster rates (so that assumption A.9 always holds).