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?



Lyapunov condition for CLT for asymptotically independent sequence
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). 


