Consider, for each $n\in\mathbb{N}$, the quantity
$$ Z_n = \frac{1}{n-1}\,\sum_{j=1}^{n-1}B_j\,B_{j+1}-\left(\frac{1}{n-1}\,\sum_{j=1}^{n-1}B_j\right)^2 $$
where $B_j$ are iid Bernoulli variables with $\mathbb{P}[B_j=1]=p$ and $\mathbb{P}[B_j=0]=1-p$. I am looking for a central limit theorem for $\sqrt{n}\,Z_n$. From simulations (I am pretty sure of the code) it seems that
$$ \sqrt{\frac{n-1}{p^2\,\left(1-p\right)^2}}\,Z_n\stackrel{d}{\longrightarrow}\text{N}(0,1)\quad (1) $$
However my computations gives me a different asymptotic variance. I made the following computations:
\begin{eqnarray} Z_n & = & \frac{1}{n-1}\,\sum_{j=1}^{n-1}(B_j\,B_{j+1}-2\,p\,B_j+p^2)-\left(\frac{1}{n-1}\,\sum_{j=1}^{n-1}(B_j-p)\right)\,\left(\frac{1}{n-1}\,\sum_{k=1}^{n-1}(B_k-p)\right). \end{eqnarray}
Now it is immediate to see that
$$ \sqrt{n-1}\,\left(\frac{1}{n-1}\,\sum_{j=1}^{n-1}(B_j-p)\right)\,\left(\frac{1}{n-1}\,\sum_{k=1}^{n-1}(B_k-p)\right)\stackrel{p}{\longrightarrow}0 $$
so that
$$ \sqrt{n-1}\,Z_n = \frac{1}{\sqrt{n-1}}\,\sum_{j=1}^{n-1}(B_j\,B_{j+1}-2\,p\,B_j+p^2)+o_p(1). $$
Now since $\mathbb{E}[B_j\,B_{j+1}-2\,p\,B_j+p^2]=0$, it follows that the variance of each $B_j\,B_{j+1}-2\,p\,B_j+p^2$ is simply
$$ \mathbb{E}\left[\left(B_j\,B_{j+1}-2\,p\,B_j+p^2\right)^2\right]=p^2\,\left(1-p^2\right), $$
hence in conclusion
$$ \sqrt{\frac{n-1}{p^2\,\left(1-p^2\right)}}\,Z_n\stackrel{d}{\longrightarrow}\text{N}\left(0,1\right) $$
which is not the result in (1) because of the $p^4$ instead of $p^2$, where am I wrong?
The error is that the random variables $P_j=B_j B_{j+1}$ are not independent: both $P_j$ and $P_{j-1}$ depend on what $B_j$ is. Moreover, they are correlated: the covariance is $p^3-p^4 \neq 0$ for $p \in (0,1)$. Thus one cannot compute the variance of their sum by simply summing the variances, you need to take into account covariances (or else compute it in a different way entirely).