Let $(a_i)_{i=1}^\infty$ be a bounded sequence of non-random constants and $(X_i)_{i=1}^\infty$ be iid Rademacher random variables (i.e. $\mathbb{P}(X_1=1)=\mathbb{P}(X_1=-1)=1/2$) such that $$\lim_{n\rightarrow\infty}\frac 1 n \sum_{i=1}^n a_i^2=\sigma^2<\infty.$$ Show that as $n\rightarrow\infty$ we have $$\frac {1}{\sqrt{n}} \sum_{i=1}^na_iX_i \xrightarrow{D}\mathcal{N}(0,\sigma^2).$$
The difficulty arises from the fact that $\{a_iX_i\}_{i=1}^\infty$ are not identically distributed so that we cannot use CLT directly. Is there some variations of CLT that I can use?
Lyapunov or Lindenbdeg CLT will do the trick, but you have some work to do to show that the sequence satisfies the condition. Let us take a look at Lindenbderg's condition: $$ \lim_{n \to \infty} \frac{\sum_{i}\mathbb{E}[a_i^2X_i^2]I\{|a_iX_i| > \epsilon s_n\}}{s_n^2} = 0. $$ Note that $|a_i X_i| = |a_i|$, and $s_n = (\sum_{i=1}^n a_i^2 )^{1/2} \ge|a_i|$, namely for any $\epsilon > 0$, $\exists n \in \mathbb{N}$ such that $|a_i| < \epsilon s_n$ as $s_n \to \infty$ (in a rate of $\sqrt{n})$, thus the condition is satisfied. Try to put it more rigorously. But this is the idea.