Lindeberg Condition for a sequence of discrete random variables. 
Let $X_1,X_2,...$ be independent and for any n $\ge 1$ and $\alpha>0$ 
  $$X_n = \left\{
\begin{array}{rl}
n^\alpha & \text{with } Pr(X_n= n^\alpha) = \frac{1}{2n^{2\alpha}},\\
-n^\alpha & \text{with }Pr(X_n= -n^\alpha) = \frac{1}{2n^{2\alpha}},\\
0 & \text{with } Pr(X_n= 0) = 1- \frac{1}{n^{2\alpha}}.
\end{array} \right.$$
  Let $S_n = X_1+ \dots +X_n$ and $B_n^2 = \sigma_1^2+\dots+\sigma_n^2.$ Does $\frac{S_n}{B_n}\rightarrow Z \sim N(0,1)$ in distribution.

Solving this question is an example of using the Lindeberg-Feller CLT. I found that,
$E[X_n]= n^\alpha(\frac{1}{2n^{2\alpha}})-n^\alpha(\frac{1}{2n^{2\alpha}})+0(1-\frac{1}{n^{2\alpha}}) = 0$
and
$E[X_n^2]=(n^{\alpha})^2(\frac{1}{2n^{2\alpha}}) + (-n^{\alpha})^2(\frac{1}{2n^{2\alpha}})+0^2(1-\frac{1}{n^{2\alpha}})=1.$
Therefore $\sigma_n^2 = 1$ and $B_n = \sqrt{n}$.
If the Lindeberg condition holds, i.e., for any $\epsilon > 0$
$$\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} E[X_k^2 I_{\{|X_k|>\epsilon B_k\}}]}{B_n^2} = 0.$$
Then $\frac{S_n}{B_n}\rightarrow Z \sim N(0,1)$ in distribution.
In our case, since $\sigma_k=1$ We have to deal with for any $\epsilon> 0,$  $\sum_{k=1}^{n} E[X_k^2 I_{\{\frac{|X_k|}{\sqrt{k}}>\epsilon\}}]$ for the numerator. I am stuck now because I dont know how to represent $E[X_k^2 I_{\{\frac{|X_k|}{\sqrt{k}}>\epsilon\}}]$.

The condition we have to check for Lindeberg's condition is thtat for all positive $\varepsilon$,
$$
\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon B_n\}}\right]}{B_n^2} = 0,
$$
and since $B_n=\sqrt n$, this is equivalent to
$$
\lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]}{n} = 0.
$$
If $\alpha\lt 1/2$, then for each fixed $\varepsilon$, there exists a $n_0$ such that $n^\alpha\leqslant \varepsilon n^{1/2}$ for all $n\geqslant n_0$ hence
for such $n$, $\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]=0$.
If $\alpha\geqslant 1/2$ and $\varepsilon\in (0,1)$, the term $\mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]$ is one if $k^\alpha\gt \varepsilon \sqrt n$, that is , if $k\gt \varepsilon^{\alpha}n^{1/(2\alpha)}$ and $0$ otherwise hence
$$
\sum_{k=1}^{n} \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]\geqslant  
\sum_{k=\lfloor \varepsilon^{\alpha}n^{1/(2\alpha)}\rfloor}^n \mathbb E\left[X_k^2 I_{\{|X_k|>\epsilon \sqrt n\}}\right]= \left(n-\lfloor \varepsilon^{\alpha}n^{1/(2\alpha)}\rfloor\right)
$$
hence Lindeberg's condition is not satistified.
Since $\max_{1 \le i \le n} \sigma_i/ s_n \to 0$, Lindeberg's condition is equivalent to the convergence of $\left(S_n/b_n\right)_n$ to a standard normal distribution. We thus conclude that the convergence of $S_n/B_n$ to a standard normal distribution holds if and only if $\alpha<1/2$.
