Let $\{X_n,\,n\geq1\}$ be independent random variables with $$\mathscr{P}\{X_k=\pm1\}=\frac{1}{2}\left(1-\frac{1}{k}\right)\mbox{ and }\mathscr{P}\{X_k=\pm\sqrt{k}\}=\frac{1}{2k}.$$Let $S_n=\sum_{k=1}^n$, we show that $S_n$ does NOT converge to normal distribution by the divergence of the Lindeberg's condition.
First, we evaluate the mean and variance of $X_k$'s and $S_n$, $$\begin{array}{rl}&\mathscr{E}(X_k)=\frac{1}{2}\left(1-\frac{1}{k}\right)-\frac{1}{2}\left(1-\frac{1}{k}\right)+\sqrt{k}\frac{1}{2k}-\sqrt{k}\frac{1}{2k}=0\\ \implies & \mathscr{E}(S_n)=\sum_{k=1}^n\mathscr{E}(X_k)=0.\end{array}$$ And, $$\sigma^2(X_k)=\mathscr{E}(X_k^2)=\frac{1}{2}\left(1-\frac{1}{k}\right)+\frac{1}{2}\left(1-\frac{1}{k}\right)+k\frac{1}{2k}+k\frac{1}{2k}=2-\frac{1}{k},$$then $$B_n^2=\sigma^2(S_n)=\sum_{k=1}^n\sigma^2(X_k)=2n-\sum_{k=1}^n\frac{1}{k}.$$Let $\eta=B_n^{-1}$, the Lindeberg's condition $$\begin{array}{rl}\frac{1}{B_n^2}\sum_{k=1}^n\mathscr{E}\left(X_k^2\,I\{|X_k|>\eta B_n\}\right) & = \frac{1}{B_n^2}\sum_{k=1}^n\mathscr{E}\left(X_k^2\,I\{|X_k|>1\}\right) \\ & = \frac{1}{B_n^2}\sum_{k=1}^n\left(k\frac{1}{2k}+k\frac{1}{2k}\right) \\ & = \frac{n}{2n-\sum_{k=1}^n\frac{1}{k}}\\ & \rightarrow \frac{1}{2}\neq0\mbox{ as }n\rightarrow\infty.\end{array}$$That is, there exists a $\eta=B_n^{-1}$ such that the Lindeberg's condition is failed. By the relationship between the Lindeberg's condition and the Lyapunov's condition, we also have that the Lyapunov's condition is failed. Thus, $S_n$ does NOT converge to normal distribution.
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