Central limit theorem for weighted random variable Let $(X_n)_n$ be a sequence of i.i.d random variable, $(x_n)_n$ be sequence of $\mathbb{R}^*.$ Let $y^2_n=\sum_{k=1}^nx^2_k.$
Suppose that $E[X_1]=0,E[X_1^2]=1,x_n=o(y_n),y_n \to +\infty.$ Prove that the sequence of the weighted random variable $(x_nX_n)_n$ fulfills the central limit theorem: $$\frac{1}{y_n}\sum_{k=1}^nx_kX_k \implies N(0;1).$$
Remark (optional): more generally, the following holds: $E[X_1^2]<+\infty$ if and only if there exist a sequence of real numbers $(w_n)_n$ such that $\frac{1}{y_n}\sum_{k=1}^nx_kX_k-w_n$ converges in distribution to an arbitrary random variable $Y.$ In this case, $Y$ is normal distributed.
A way to prove that the CLT is fulfilled, is to prove that Lindeberg condition holds. So let $\epsilon>0.$
$$\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_k^2 1_{|x_kX_k|>\epsilon y_n}]=\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_1^2 1_{|x_kX_1|>\epsilon y_n}],$$ can't see how to continue from here, especially how to remove $x_k$ from $1_{|x_kX_1|>\epsilon y_n}?$
Any suggestions are welcomed.

Part 1:
Counterexample.
Consider $X_k$ -i.i.d., $X_k \sim U[-1,1]$, $x_1 = 1$, $x_k  = \frac1{2^{k+2}}$, $k \ge 2$.
$$S_n = \frac{ X_1 + \eta_n}{ \sqrt{(1 + \sum_{k=1}^n  \frac1{4^{k+2}}) }}$$
where $|\eta_n| = |\sum_{k = 2}^n \frac1{2^{k+2}} X_k| \le \sum_{k = 2}^n \frac1{2^{k+2}} \le \frac12$.
As $\eta_n = \sum_{k = 2}^n \frac1{2^{k+2}} X_k $ we have $\exists \eta:$
$\eta_n \to \eta$ a.s. (Kolmogorov's two-series theorem). As$|\eta_n| \le \frac12$ we have $\eta \le \frac12$.
Put $c = \sqrt{1 + \sum_{k=1}^{\infty}  \frac1{4^{k+2}} }$. Hence
$$S_n \to \frac{X_1 + \eta}{c}  $$ a.s. But $|\frac{X_1 + \eta}{c} | \le \frac{2}c$ and hence
$$\lim_n S_n =   \frac{X_1 + \eta}{c} \ne N(0,1).$$
The claim is false. Hence we need some stronger assumption, and assumption $\max_{1 \le k \le n}|x_k| = o(y_n)$ is sufficient, as it's shown below.
Part 2: Suppose that we don't have condition "$X_n$ are i.i.d." - in this case problem is even more interesting.
Claim: instead of condition $x_n = o(y_n)$ we should have more strong condition $\max_{1 \le k \le n}|x_k| = o(y_n)$, otherwise there's no convergence to $N(0,1)$. Let us prove it.
Theorem (from Probability 1 by A.N. Shiryaev): if $\xi_{n1}, \ldots,  \xi_{nn}$ is a sequence of independent r.v., $E \xi_{nk}=0$, $DS_n = 1$, where $S_n = \xi_{n1} + \ldots + \xi_{nn}$ then $S_n \to N(0,1)$ iff $\max_{1 \le k \le n}E\xi_{kn}^2 \to 0$, $n \to \infty$.
Put $\xi_{nk} = \frac{x_k X_k}{y_n}$. We have $E \xi_{nk}=0$, $DS_n = 1$. So $S_n \to N(0,1)$ iff
$$\max_{1 \le k \le n}E\xi_{kn}^2 \to 0, n \to \infty.$$
As $\max_{1 \le k \le n}E\xi_{kn}^2 = \max_{1\le k\le n}\frac{x_k^2}{y_n^2}$ we have that $S_n \to N(0,1)$ iff $max_{1 \le k \le n}|x_k| = o(y_n)$.
For example, we have convergence to $N(0,1)$ for  $x_n$ such that $|x_n|$ is nondecreasing, but not in general case.
Сonclusion: if $X_i$ are not necessary i.i.d., then the condition $\max_{1 \le k \le n}|x_k| = o(y_n)$ is necessary and sufficient  to guarantee the convergence to $N(0,1)$ for all independent $X_i$ with $EX_i = 0$ and $DX_i = 1$.
Let us notice also, that if $X_i \sim N(0,1)$ then $S_n = N(0,1)$  for any $x_k$.
