Let independent $X_{ni}\sim BIN(1,\theta_{ni}), 1\leq i \leq n$ and suppose $\sum_{i=1}^{n}\theta_{ni}(1-\theta_{ni})\to \infty$.
Show that $$\frac{\sum_{i=1}^{n}X_{ni}-\sum_{i=1}^{n}\theta_{ni}}{\sqrt{\sum_{i=1}^{n}\theta_{ni}(1-\theta_{ni})}}\to N(0,1)$$(converge in law)
Since $M_{X_{ni}}(t)=\theta_{ni}e^t+1-\theta_{ni}$ then $$ M_{X_{ni}-\theta_{ni}}(t)=\left(\theta_{ni}e^t+1-\theta_{ni}\right)e^{-t\theta_{ni}} $$ and $$ M_{\sum_{i=1}^n(X_{ni}-\theta_{ni})}(t)=\prod_{i=1}^n\left(\theta_{ni}e^t+1-\theta_{ni}\right)e^{-t\theta_{ni}} $$ Denote the sum in denominator by $N=N(n)$ for shortness: $N=\sum_{i=1}^n\theta_{ni}(1-\theta_{ni})$. Then $$ M_{\frac{\sum_{i=1}^n(X_{ni}-\theta_{ni})}{\sqrt{N}}}(t)=\prod_{i=1}^n\left(\theta_{ni}e^{\frac{t}{\sqrt{N}}}+1-\theta_{ni}\right)e^{-\frac{t}{\sqrt{N}}\theta_{ni}} = \prod_{i=1}^n\left(\theta_{ni}e^{\frac{t}{\sqrt{N}}(1-\theta_{ni})}+(1-\theta_{ni})e^{-\frac{t}{\sqrt{N}}\theta_{ni}}\right)$$ Use Taylor expansions of exponent as $N(n)\to\infty$ for $n\to\infty$: $$ e^{\frac{t}{\sqrt{N}}(1-\theta_{ni})} = 1+\frac{t}{\sqrt{N}}(1-\theta_{ni})+\frac{t^2}{2N}(1-\theta_{ni})^2+\frac{t^3}{6N^{3/2}}(1-\theta_{ni})^3+o\left(\frac{t^3}{6N^{3/2}}(1-\theta_{ni})^3\right) $$ and $$ e^{-\frac{t}{\sqrt{N}}\theta_{ni}} = 1-\frac{t}{\sqrt{N}}\theta_{ni}+\frac{t^2}{2N}\theta_{ni}^2 - \frac{t^3}{6N^{3/2}}\theta_{ni}^3 +o\left(\frac{t^3}{6N^{3/2}}\theta_{ni}^3\right) $$ I use expansion to the third term just in case, suddenly the second term will not be enough. Substitute exponents in the product by expansions: $$ M_{\frac{\sum_{i=1}^n(X_{ni}-\theta_{ni})}{\sqrt{N}}}(t)=\prod_{i=1}^n\biggl(1+\frac{t^2}{2N}\theta_{ni}(1-\theta_{ni})+\underbrace{\frac{t^3}{6N^{3/2}}\theta_{ni}(1-\theta_{ni})(\theta_{ni}^2+(1-\theta_{ni})^2)}_{\alpha_{ni}}+o(\alpha_{ni}) \biggr) $$ Then take $\ln$ and expand $\ln(1+x)=x-\frac{x^2}{2}+o\left(\frac{x^2}{2}\right)$ as $x\to 0$. $$ \ln M = \frac{t^2}{2N}\underbrace{\sum_{i=1}^n \theta_{ni}(1-\theta_{ni})}_N + \sum_{i=1}^n \alpha_{ni} + o\left(\sum_{i=1}^n \alpha_{ni}\right). $$ All the other terms are included in $o(\ldots)$ since there are $N^2$ and even more in their denominators.
Consider $\sum_{i=1}^n \alpha_{ni}$: since $\theta_{ni}^2+(1-\theta_{ni})^2\leq 1$, $$ \sum_{i=1}^n \alpha_{ni} \leq \sum_{i=1}^n \frac{t^3}{6N^{3/2}}\theta_{ni}(1-\theta_{ni}) =\frac{t^3}{6N^{3/2}}\cdot N = \frac{t^3}{6\sqrt{N}} \to 0 $$ Therefore $$ \ln M \to \frac{t^2}{2} \text{ as } n\to\infty $$