Given $X_1,X_2,\ldots$ – i.i.d, with $\mathbb{E}X_1=0$, $0<\mathbb{D}X_1<\infty$, and $S_n=\sum_{i=1}^n X_i$, one must prove that $$ \forall \epsilon > 0:\ \lim_{n\rightarrow \infty}P\left(\left|\frac{S_n}{n^\alpha}\right|\leq \epsilon\right) = \begin{cases} 0,&\alpha<\frac{1}{2}\\ 1,&\alpha>\frac{1}{2} \end{cases} .$$
I know how to prove the case where $\alpha>\frac{1}{2}$, namely by applying the Chebyshev inequality. However, what can I do for the case where $\alpha<\frac{1}{2}$?
The central limit limit theorem says that $$\lim_{N\to \infty} \mathbb{P}\left(\left|\frac{S_n}{\sigma n^{1/2}}\right|\le \epsilon\right)=\frac{1}{\sqrt{2\pi}}\int_{|x|\le \epsilon}e^{-x^2/2}dx=\mathbb{P}\left(\left|\cal{N}\right|\le \epsilon\right).$$
Now, if $\alpha<1/2$, we may choose, for any $M>0$, some $N$ such that for $n>N$, $1/n^{\alpha}>M/\sigma n^{1/2}$. For such $n$, we thus have that $$\mathbb{P}\left(\left|\frac{S_n}{n^{\alpha}}\right|\le \epsilon\right)\le \mathbb{P}\left(\left|\frac{S_n}{\sigma n^{1/2}}\right|\le \frac{\epsilon}{M}\right).$$ Taking limits of both sides, we get that $$\lim_{n\to \infty}\mathbb{P}\left(\left|\frac{S_n}{n^{\alpha}}\right|\le \epsilon\right)\le \mathbb{P}\left(\left|\mathcal{N}\right|\le \frac{\epsilon}{M}\right).$$ As $M$ was arbitrary, we may also take the limit as $M\to \infty $ on both sides to get that $$\lim_{n\to \infty}\mathbb{P}\left(\left|\frac{S_n}{n^{\alpha}}\right|\le \epsilon\right)\le \lim_{M\to \infty}\mathbb{P}\left(\left|\mathcal{N}\right|\le \frac{\epsilon}{M}\right)=0.$$