In Politis et al. (1997)'s proof of Theorem A.1 (specifically, pages 309-310), there are issues that I'm trying to clarify. In this previous post, it is clear that the following equality does not hold:
With the correct powers, the expression does not converges to zero (and the proof breaks down). Now, take a look at the sketch of the proof:
As far as I understand, Step $1$ should be the Lyapunov's condition. I will state the Lyapunov's Central Limit Theorem.
Suppose that $\{X_i\}$ is a sequence of independent random variables such that, for each $i$, $E(X_i)=\mu_i<\infty$ and $ Var(X_i)=\sigma_i^2<\infty$. Define $s_n^2=\sum_{k=1}^n \sigma_k^2$. If there exists $\delta>0$ so that $\mid X_i\mid^{2+\delta}$ are integrable and the Lyaponov's condition holds, i.e., $$\lim_{n\to\infty}\frac{1}{s_n^{2+\delta}}\sum_{k=1}^n E[\mid X_k-\mu_k\mid^{2+\delta}]=0,$$ then $$\frac{1}{s_n}\sum_{k=1}^n (X_k-\mu_k)\overset{d}{\to} N(0,1).$$
Assume that $\{U'_{n,i}\}_{i=1}^{r_n}$ are zero mean random variables satisfying the moment conditions of the theorem. I believe that Step $1$ should be replaced by $$\bigg[Var\bigg(\sum_{k=1}^{r_n} U'_{n,i}\bigg)\bigg]^{-(2+\delta)/2}\sum_{i=1}^{r_n} E\mid U'_{n,i}\mid^{2+\delta}\to 0, n\to\infty$$ since $\big[Var\big(\sum_{k=1}^{r_n} U'_{n,i}\big)\big]^{-(2+\delta)/2}=\big[\sum_{k=1}^{r_n} Var(U'_{n,i})\big]^{-(2+\delta)/2}:=\big[s_n^2\big]^{-(2+\delta)/2}=1/s_n^{2+\delta}$. Doing so, I can easily show the convergence of the term of Step $1$, using exactly the same powers that the author used.
To conclude, I believe the authors typed the power of the variance term wrong.
Do you agree with me?