$Z_1, Z_2,...$ are iid uniformly distributed on $[-1;1]$,
$\lim_{n \to \infty} a_n=0$ and $\lim_{n \to \infty} na_n=\infty$ also $a_n>0$ $\forall n$,
$X_{n,j}= \frac{1}{a_n}I(|Z_j| \le a_n)$ $\forall j = 1,...,n$ where $I$ is the indicator function,
$S_n=\sum_{j=1}^{n}X_{n,j}$.
One needs to show that $(S_n-n)\sqrt{\frac{a_n}{n}} \to N(0,1)$ in distribution.
Easy to show that:
$E[S_n]=n$ if $n$ large enough,
also $Var[S_n]= \frac{(1-a_n)n}{a_n}$ if $n$ large enough,
and the Lyapunov condition is satisfied for $\delta=2$, i.e. $$\lim_{n \to \infty} \sum_{j=1}^{n} \frac{1}{Var(S_n)^2}E[|X_{n,j}-E(X_{n,j})|^4]=0.$$
But then it follows that $\frac{S_n-E[S_n]}{\sqrt{Var(S_n)}}=(S_n-n)\sqrt{\frac{a_n}{n(1-a_n)}} \to N(0,1)$ in distribution. What am I doing wrong?
Reaction to Did's comment:
$Z_n=(S_n-n)\sqrt{\frac{a_n}{n(1-a_n)}}$ and $Y_n=Z_n*\sqrt{(1-a_n)}$, Clearly if $F_{Z_n}(x)$ is a cdf for $Z_n$, then $F_{Z_n}(\frac{x}{\sqrt{1-a_n}})$ is the cdf for $Y_n$. Now we need to prove that $\lim_{n \to \infty}F_{Z_n}(\frac{x}{\sqrt{1-a_n}})=\Phi(x)$ $\forall x$. And here I am stuck.