Proving that the law of large numbers doesn't apply 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 $\alpha<\frac{1}{2}$?
 A: 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
$$\lim_{n\to \infty}\mathbb{P}\left(\left|\frac{S_n}{n^{\alpha}}\right|\le \epsilon\right)\le \lim_{M\to \infty}\mathbb{P}\left(\left|\mathcal{N}\right|\le \frac{\epsilon}{M}\right)=0.$$
