The sequence of independent random variables has the following distribution.
$$
\alpha>0\,, \quad P(X_n=n)=P(X_n=-n)=\frac{1}{6n^{2(\alpha-1)}}\,,\quad P(X_n=0)=1-\frac{1}{3n^{2(\alpha-1)}}
$$
I need to prove that the Lindeberg condition holds if and only if $\alpha<\frac{3}{2}$.
In the case of above distribution, I turn the problem into proving
$$
\frac{\sum_{k=\min\left(a:a\geq \epsilon \sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}\right)}^{n}\frac{1}{3}k^{4-2\alpha}}{\sum_{k=1}^{n}\frac{1}{3}k^{4-\alpha}} \rightarrow 0
$$
How to get the result?
Lindeberg condition:
$$
\{ X_n,n\ge1 \} \text{is a sequence of independent random variable.}\\EX_k=\mu_k \quad \operatorname{Var}X_k=\sigma_k^2 \quad
B_n=\sum_{k=1}^{n} \sigma_k^2 \\ \forall \epsilon > 0 \quad \frac{1}{B_n}\sum_{k=1}^{n}\int_{|x-\mu_k|>\epsilon\sqrt{B_n}}(x-\mu_k)^2dF_k(x) \rightarrow 0 \quad (n\rightarrow \infty)
$$
Best Answer
We have to check that $$R_n:=\frac{\sum_{k=\min\left(a:a\geq \epsilon \sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}\right)}^{n} k^{4-2\alpha}}{\sum_{k=1}^{n} k^{4-2\alpha}} \rightarrow 0.$$ Let $B_n=\sum_{i=1}^{n}\operatorname{Var}(X_i)$. If $\alpha\gt 5/2$, then $B_n$ converges to a positive number and $R_n$ will converge to $1$.
If $\alpha =5/2$, then $R_n$ behaves as $$ R'_n:=\frac 1{\log n}\sum_{k=\lfloor c \log n\rfloor }^nk^{-1} $$ which goes also to $1$.
Now, assume that $\alpha\lt 5/2.$ Then one can find constants $c_1$ and $c_2$ depending only on $\alpha$ such that $c_1n^{5-2\alpha}\leqslant B_n\leqslant c_2n^{5-2\alpha}$ and a similar bound holds for $B_n$. As a consequence, the summation set in the denominator is empty for $n$ large enough in the case $\alpha\lt 3/2$.
If $\alpha \gt 3/2$, one gets
$$ \sum_{k=\min\left(a:a\geq \epsilon \sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}\right)}^{n} k^{4-2\alpha}\geqslant \sum_{k=\lfloor Kn\rfloor}^nk^{4-2\alpha}\geqslant K'n^{5-2\alpha} $$ for some positive $K'$ hence $R_n$ does not go to zero.