Application of Lindeberg's Central Limit Theorem (2)

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Let $X_j$ be defined as follows for some $\alpha>1$: $$X_j=\begin{cases}\pm j^\alpha, &\mbox{ with probability }\frac{1}{6\,j^{2(\alpha-1)}}\mbox{ each;}\\0,&\mbox{ with probability }1-\frac{1}{3\,j^{2(\alpha-1)}}.\end{cases}$$ We have $$\begin{array}{rl}
\mathscr{E}(X_j) &=0. \\
\sigma^2(X_j) & =\mathscr{E}(X_j^2)=j^{2\alpha}\frac{2}{6\,j^{2(\alpha-1)}}=\frac{j^2}{3}\implies \\
\sigma^2_n &=\sigma^2(S_n)=\sum_{j=1}^n\sigma^2(X_j)=\sum_{j=1}^n\frac{j^2}{3}=\frac{n(n+1)(2n+1)}{18}. \\
\end{array}$$The Lindeberg's condition is defined as, let $\eta>0$, $$\begin{array}{rl}
\frac{1}{\sigma^2_n}\sum_{j=1}^n\mathscr{E}\left(X_j^2\,I\{|X_j|>\eta\sigma_n\}\right)
& = \frac{1}{\sigma^2_n}\sum_{j=1}^n\mathscr{E}\left(X_j^2\,I\{j^\alpha>\eta\sigma_n\}\right) \\
(\because\,j\leq n)& \leq \frac{1}{\sigma^2_n}\sum_{j=1}^n\mathscr{E}\left(X_j^2\,I\{n^\alpha>\eta\sigma_n\}\right) \\
 & = \frac{1}{\sigma^2_n}\left[\sum_{j=1}^n\mathscr{E}(X_j^2)\right]\,I\{n^\alpha>\eta\sigma_n\} \\
 & = I\{n^\alpha>\eta\sigma_n\}. \\
\end{array}$$Hence, when $I\{n^\alpha>\eta\sigma_n\}=0$, Lindeberg's condition holds.  We have $$I\{n^\alpha>\eta\sigma_n\}=0 \iff n^\alpha<\eta\sigma_n \iff n^\alpha=o(n^{\frac{3}{2}}) \iff \alpha<\frac{3}{2}.$$The Lindeberg's condition is satisfied if and only if $\alpha<3/2$.