Let $\{X_n,\,n\geq1\}$ be independent random variables with $$\mathscr{P}\{X_k=\pm1\}=\frac{1}{2}\left(1-\frac{1}{k^2}\right)\mbox{ and }\mathscr{P}\{X_k=\pm\sqrt{k}\}=\frac{1}{4k^2}.$$Let $S_n=\sum_{k=1}^n$, we show that $S_n/B_n$ converges to normal distribution for some normalizing constants $B_n$.
First, we evaluate the mean and variance of $X_k$'s and $S_n$, $$\begin{array}{rl}&\mathscr{E}(X_k)=\frac{1}{2}\left(1-\frac{1}{k^2}\right)-\frac{1}{2}\left(1-\frac{1}{k^2}\right)+\sqrt{k}\frac{1}{4k^2}-\sqrt{k}\frac{1}{4k^2}=0\\ \implies & \mathscr{E}(S_n)=\sum_{k=1}^n\mathscr{E}(X_k)=0.\end{array}$$ And, $$\sigma^2(X_k)=\mathscr{E}(X_k^2)=\frac{1}{2}\left(1-\frac{1}{k^2}\right)+\frac{1}{2}\left(1-\frac{1}{k^2}\right)+k\frac{1}{4k^2}+k\frac{1}{4k^2}=1+\frac{1}{2k}-\frac{1}{k^2},$$then $$B_n^2=\sigma^2(S_n)=\sum_{k=1}^n\sigma^2(X_k)=n+\sum_{k=1}^n\frac{1}{2k}-\sum_{k=1}^n\frac{1}{k^2}.$$Let $\delta=2$ in the Lyapunov's condition, we have $$\mathscr{E}(X_k^4)=\frac{1}{2}\left(1-\frac{1}{k^2}\right)+\frac{1}{2}\left(1-\frac{1}{k^2}\right)+k^2\frac{1}{4k^2}+k^2\frac{1}{4k^2}=\frac{3}{2}-\frac{1}{k^2},$$thus the Lyapunov's condition $$\frac{1}{B_n^4}\sum_{k=1}^n\mathscr{E}(X_k^4)=\frac{\frac{3n}{2}-\sum_{k=1}^n\frac{1}{k^2}}{\left[n+\sum_{k=1}^n\frac{1}{2k}-\sum_{k=1}^n\frac{1}{k^2}\right]^2}\rightarrow0\mbox{ as }n\rightarrow\infty,$$holds by comparing the order of $n$. Hence we have $$\frac{S_n}{B_n}\overset{d}{\rightarrow}\Phi.$$
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