Convergence of the sum of iid scaled by $n^\alpha$

Question

I am interested in the convergence of the sequence $\mathbb{P}(|X_1+...+X_n|/n^\alpha<z)$ where $z>0$, $\{X_n\}_n$ is an i.i.d. sequence with mean zero and finite variance. I can easily prove that the sequence converges to 1 when $\alpha>1/2$ using Chebyshev's inequality. However I am struggling to show what happens when $\alpha<1/2$. Intuitively it should go to zero, any help would be appreciated.

Answer 1

The central limit theorem combined with the continuous mapping theorem says that $$ \frac{1}{n^{1/2}}\left\lvert \sum_{i=1}^nX_i\right\rvert\to \left\lvert N\right\rvert, $$ where $N$ has a centered normal distribution.

When $\alpha <1/2$, we get that $$ \mathbb P\left( \frac{1}{n^{\alpha}}\left\lvert \sum_{i=1}^nX_i\right\rvert\lt z\right)=\mathbb P\left( \frac{1}{n^{1/2}}\left\lvert \sum_{i=1}^nX_i\right\rvert\lt n^{\alpha-1/2}z \right). $$ Fix a positive $\varepsilon$ which is a continuity point of the cumulative distribution function of $\left\lvert N\right\rvert$. For a fixed $z$ and $n$ large enough, the term $n^{\alpha-1/2}z$ is smaller than $\varepsilon$ hence for these $n$, $$ \mathbb P\left( \frac{1}{n^{\alpha}}\left\lvert \sum_{i=1}^nX_i\right\rvert\lt z\right)\leqslant \mathbb P\left( \frac{1}{n^{1/2}}\left\lvert \sum_{i=1}^nX_i\right\rvert\lt\varepsilon \right) $$ and taking the $\limsup_{n\to +\infty}$ gives $$ \limsup_{n\to +\infty}\mathbb P\left( \frac{1}{n^{\alpha}}\left\lvert \sum_{i=1}^nX_i\right\rvert\lt z\right)\leqslant \mathbb P\left( \left\lvert N\right\rvert\lt\varepsilon \right). $$