Lyapunov condition for CLT for asymptotically independent sequence

Suppose I have some triangular array $$\{X_{n,j}\}$$ of random variables, which need not be independent or identically distributed. Suppose I further know that $$Var\left(\sum_{j=1}^n X_{n,j}\right)\to \sigma^2 \text{ and } \sum_{j=1}^n \mathbb{E}|X_{n,j}|^{2+\epsilon}\to 0,$$ for some $$\epsilon>0$$. Then this triangular array satisfies the Lyapunov condition. However, for Lyapunov's CLT we further require independence for each $$n$$. Suppose instead that we have asymptotic independence in the sense that for any $$t\in\mathbb{R},$$ $$\left|\mathbb{E}\left[\prod_{j=1}^n\exp\left(itX_{n,j}\right)\right] - \prod_{j=1}^n\mathbb{E}\left[\exp\left(itX_{n,j}\right)\right]\right|\to 0,$$ where $$i$$ denotes the imaginary unit. Can we then still conclude that $$\sum_{j=1}^n X_{n,j}\to N(0, \sigma^2)?$$

It is easy to see (by a Levy continuity theorem argument) that this is true if the covariances also fade fast enough, so that $$\sum_{j=1}^nVar\left( X_{n,j}\right)\to \sigma^2,$$ but do we need this or can we also conclude asymptotic normality without this additional condition?

The answer is no. E.g., suppose that $$\begin{equation*} (X_{n,j})=(X_{n,j})_{j=1}^n\sim(1-u^2/n)N_n(0,I_n/n)+(u^2/n) N_n(0,J_n/n), \end{equation*}$$ where $$u\in(0,\infty)$$, $$I_n$$ is the $$n\times n$$ identity matrix, and $$J_n$$ is the $$n\times n$$ matrix with all entries equal $$1$$, so that $$N_n(0,J_n/n)$$ is the distribution of the random vector $$(Y,\dots,Y)/\sqrt n$$ with $$Y\sim N(0,1)$$. Thus, the distribution of $$(X_{n,j})$$ is the mixture of the $$n$$-variate normal distributions $$N_n(0,I_n/n)$$ and $$N_n(0,J_n/n)$$ with respective weights $$1-u^2/n$$ and $$u^2/n$$.
The idea here is to make the variance of the sum $$\sum_j X_{n,j}$$ greater (than it would be with $$u=0$$) without significantly affecting the distribution of the sum, which will still be $$\approx N(0,1)$$ even if $$u\ne0$$.
Then $$\begin{equation*} Cov(X_{n,j},X_{n,k})=(1-u^2/n)1_{j=k}/n+(u^2/n)/n \end{equation*}$$ and hence $$\begin{equation*} Var\sum_jX_{n,j}=\sum_{j,k}Cov(X_{n,j},X_{n,k}) \\ =(1-u^2/n)+n(n-1)(u^2/n)/n\to1+u^2=:\sigma^2. \tag{1} \end{equation*}$$ Next, for any real $$\epsilon>0$$ $$\begin{equation*} E|X_{n,j}|^{2+\epsilon}=(1-u^2/n)/n^{1+\epsilon/2}+(u^2/n)/n^{1+\epsilon/2} =1/n^{1+\epsilon/2} \end{equation*}$$ and hence $$\begin{equation*} \sum_jE|X_{n,j}|^{2+\epsilon}=n/n^{1+\epsilon/2}\to0. \end{equation*}$$ Similarly, $$Ee^{itX_{n,j}}=e^{-t^2/(2n)}$$ and hence $$\begin{equation*} \prod_1^n Ee^{itX_{n,j}}=e^{-t^2/2}. \end{equation*}$$ Further, $$\begin{equation*} E\prod_1^n e^{itX_{n,j}}=(1-u^2/n)e^{-t^2/2}+O(u^2/n)\to e^{-t^2/2}, \end{equation*}$$ so that the condition $$\begin{equation*} E\prod_1^n e^{itX_{n,j}}-\prod_1^n Ee^{itX_{n,j}}\to0 \end{equation*}$$ holds as well. However, $$\begin{equation*} \sum_j X_{n,j}\sim(1-u^2/n)N(0,1)+(u^2/n) N(0,n)\to N(0,1)\ne N(0,\sigma^2), \end{equation*}$$ in view of (1).