Let $\{X_{i}\}_{i\geq1}$ be a collection of independent r.v's with probabilities $$\mathbb{P}(X_{n}=n^{\alpha})=\mathbb{P}(X_{n}=-n^{\alpha})=\frac{1}{2}$$ Show that the CLT is valid as long as $\alpha > -\frac{1}{2}$.
We know that $\mathbb{E}(X_{n}) = 0, \quad \mathbb{E}(X_{n}^{2}) = n^{2\alpha}, \quad\mathbb{E}(|X_{n}|^{2+\delta}) = n^{\alpha(2+\delta)}$.
By the Liapunov condition we get the following:
$$\frac{\sum_{k=1}^{n}\mathbb{E}(|X_{k}-\mu_{k}|^{2+\delta})}{((\mathbb{Var}X_{n})^{\frac{1}{2}})^{2+\delta}}=\frac{\sum_{k=1}^{n}k^{\alpha(2+\delta)}}{\left(\sum_{k=1}^{n}k^{2\alpha}\right)^{\frac{2+\delta}{2}}}$$ Why can't $\alpha < -\frac{1}{2}$? I don't see it clearly, sorry if it is a simple question and thanks for the help.
For $n\geqslant 2$ and $p\gt -1$, the following inequalities hold: $$ c_pn^{p+1} \leqslant\sum_{k=\lfloor n/2\rfloor+1}^n k^p\leqslant \sum_{k=1}^n k^p\leqslant C_p n^{p+1} $$ hence for $\alpha>-1/2$, there exists constants $C_\delta$ and $C'_\delta$ such that $$ C_\delta n^{\alpha(2+\delta)+1-\frac{2+\delta}2\left(2\alpha+1\right)}\leqslant \frac{\sum_{k=1}^{n}\mathbb{E}(|X_{k}-\mu_{k}|^{2+\delta})}{((\mathbb{Var}X_{n})^{\frac{1}{2}})^{2+\delta}}\leqslant C'_\delta n^{\alpha(2+\delta)+1-\frac{2+\delta}2\left(2\alpha+1\right)}. $$ Since $$ \alpha(2+\delta)+1-\frac{2+\delta}2\left(2\alpha+1\right)=-\delta/2, $$ Lyapunov's condition holds.
For $\alpha\leqslant -1/2$, this does not apply since the series $\sum_{k=1}^{+\infty}k^{2\alpha}$ converges.