Sum converging to a Normal Say $(X_i)$ is a sequence of $n$ independent Bernoulli random variables, with parameters $p_i$,  $i:1...n$. How do I prove that the random variables,
$$
S_{n} =\frac{1}{n} \sum _{i=1}^{n}a_iX_i
$$
converge in distribution to a normal as $n\to \infty$, where $0<a_i<1$ are real constants? Note: The classic CLT requires a simple sum of $i.i.d$ random variables... I think we need the Lyapunov version, but I'm not sure how to check the conditions.
 A: Let $e_n=E(S_n)=\frac{1}{n}\sum_{i=1}^na_i.p_i$
let $\sigma_n=\sigma(S_n)=\frac{1}{n}\sqrt{\sum_{i=1}^na_i^2p_i(1-p_i)}$
$$L_n=\frac{S_n-e_n}{\sigma_n}$$
(We suppose to have $0<p_1<1$ to obtain $\sigma_n>0$)
Hence, $E(L_n)=0$ and $V(L_n)=1$. We compute the characteristic function of $L_n$ :
$$\phi_{L_n}(t)=e^{i\frac{-e_n}{\sigma_n}t}\prod_{i=1}^n(1-p_i+p_ie^{i\frac{a_i}{n.\sigma_n}t}) $$
We use the fact the $e^{\epsilon}\approx 1+\epsilon+\frac{\epsilon^2}{2}$ and we suppose that $\lim(n.\sigma_n)=\infty$. (we NEED it !)
$$\lim(\phi_{L_n}(t))\approx e^{i\frac{-e_n}{\sigma_n}t}\prod_{i=1}^n(1+p_i.i\frac{a_i}{n.\sigma_n}t-\frac{p_i}{2}\left(\frac{a_i}{n.\sigma_n}t\right)^2)\approx_* e^{i\frac{-e_n}{\sigma_n}t}. e^{i\frac{e_n}{\sigma_n}t}.e^{-\frac{t^2}{2}}=e^{-\frac{t^2}{2}} $$
Hence, this is convergent to a normal law.
for (*) you need also $\ln(1+\epsilon)\approx \epsilon-\frac{\epsilon^2}{2}$
More generally, you can see from the proof that you need $$\lim (\forall i)\frac{a_i}{n\sigma_n}=0$$
