Theorem 5.2 in Srivastava(2005) is about the distribution of a test statistic for testing whether the covariance matrix is diagonal or not. In the paper they gave the following result without showing proof: For $$\gamma^*=\frac{tr(S^2)-\frac{1}{n}tr(S)^2}{\sum_{i=1}^{p}s_{ii}^2}=\frac{a_2^*}{a_{20}^*}$$
They state that $$\gamma^* \rightarrow N(\gamma-1,(a_{20}^2)^{-1}\frac{4}{n^2}(a_{2}^2-p^{-1}a_4))$$ where $S$ is the sample covariance matrix, $\Sigma$ is the population covariance matrix. $i,j$ element of S is denoted as $s_{ij}$. $a_2=tr(\Sigma^2)/p$, $a_4=tr(\Sigma^4)/p$, $a_{20}=tr(diag(\Sigma)^2)/p$. And $a_2^*=(tr(S^2)-\frac{1}{n}tr(S)^2)/p$ and $a_{20}^*=tr(diag(S)^2)/p$.
I don't understand why they could let the denominator converge to it's mean when both the numerator and denominator are $O_p(\frac{1}{\sqrt(np)})$. They also showed $a_2^*$ can be decomposed into two parts $q1$ and $q2$, the former is $O_p(\frac{1}{\sqrt(np)})$ and the latter is $O_p(\frac{1}{n})$ But the variance term is $Var(q_2)/{a_{20}^2}$ and the assumption on the relationship between $p$ and $n$ is $n=O_p(p^\delta)$ where $0<\delta\leq1$. So how can it be true when $\delta=1$?
I'm quite confused as it's a well-known theorem in sphericity tests so I hope someone could enlighten me on how this is derived. Thanks!
The link for the paper is here: https://www.jstage.jst.go.jp/article/jjss/35/2/35_2_251/_pdf/-char/ja