Lindeberg CLT application Let $X_1,X_2,\dots$ be a sequence of independent Random Variables with
$$\mathbb{P}(X_k = -1) = \mathbb{P}(X_k = 1) = \frac{1 - 2^k}{2}$$
$$\mathbb{P}(X_k = 2^k) = \mathbb{P}(X_k = -2^k) = \frac{1}{2^{k+1}} $$

How can I show the CLT holds? Tried to verify Lindeberg's Condition without success.
First, use Borel-Cantelli Lemma to show that $\mathbb{P}[|X_k|&gt;1,~\text{for some}~k\geq N]\leq 1/2^{N-1}$. Then, use the distributions of $\frac{S_n}{\sqrt{n}}$ and $\frac{\tilde{S_n}}{\sqrt{n}}$, and show the difference between them. Here $\tilde{S_n}=\sum_{i=1}^n\tilde{X_i}$ and $\tilde{X_k}$ satisfies $\mathbb{P}[\tilde{X_k}=1]=\mathbb{P}[\tilde{X_k}=-1]=1/2$.

There are three steps to show that.

Step 1, for any given $N$, there exists a $M&gt;N$, such that $\frac{|S_n-\bar{S_n}|}{\sqrt{n}}&lt;1/2^{N-1}$ for any $n&gt;M$ while $\omega\in\Omega\backslash\mathcal{A}_N$. Here $\mathcal{A}_N\triangleq\{|X_k|&gt;1 ~\text{for some}~k\geq N\}$.

Step 2, notice that $\mathbb{P}[\mathcal{A}_N]\leq 1/2^{N-1}$. Then you get $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\leq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x+1/2^{N-1}]+1/2^{N-1}$, as well as $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\geq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x-1/2^{N-1}]-1/2^{N-1}$.

Step 3, let $n$ tend $\infty$, and notice that $N$ can be arbitrary natural number. Then you can obtain the conclusion.
First, use Borel-Cantelli Lemma to show that $\mathbb{P}[|X_k|&gt;1,~\text{for some}~k\geq N]\leq 1/2^{N-1}$. Then, use the distributions of $\frac{S_n}{\sqrt{n}}$ and $\frac{\tilde{S_n}}{\sqrt{n}}$, and show the difference between them. Here $\tilde{S_n}=\sum_{i=1}^n\tilde{X_i}$ and $\tilde{X_k}$ satisfies $\mathbb{P}[\tilde{X_k}=1]=\mathbb{P}[\tilde{X_k}=-1]=1/2$.

There are three steps to show that.

Step 1, for any given $N$, there exists a $M&gt;N$, such that $\frac{|S_n-\bar{S_n}|}{\sqrt{n}}&lt;1/2^{N-1}$ for any $n&gt;M$ while $\omega\in\Omega\backslash\mathcal{A}_N$. Here $\mathcal{A}_N\triangleq\{|X_k|&gt;1 ~\text{for some}~k\geq N\}$.

Step 2, notice that $\mathbb{P}[\mathcal{A}_N]\leq 1/2^{N-1}$. Then you get $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\leq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x+1/2^{N-1}]+1/2^{N-1}$, as well as $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\geq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x-1/2^{N-1}]-1/2^{N-1}$.

Step 3, let $n$ tend $\infty$, and notice that $N$ can be arbitrary natural number. Then you can obtain the conclusion.