Let $(X_i)_{i\geqslant 1}$ be a sequence of i.i.d. random variables in $\mathcal{L}^2$ with $\mathbb{E}(X_i)=\mu$, and finite standard deviation $\sigma>0$. Let $S_n:=\sum_{i=1}^n X_i$.
The Central Limit Theorem tells that $$ Z_n:=\frac{S_n-n\mu}{\sigma\sqrt{n}}\to\mathcal{N}(0,1). $$
That is, $$ \lim_{n\to\infty}P(Z_n\leq z)=\Phi(z). $$
Now, I've read that this implies that $$ P(S_n\geq n\mu+n^{\alpha})\to 0\text{ whenever }\alpha>\frac{1}{2}. $$
How does this follow?
You can rewrite $S_n$ in terms of $Z_n$ by rearranging the expression given in the CLT, so that, $$S_n=\sigma\sqrt{n}Z_n +\mu n $$ and replacing this in the expression at the end, you get, $$P(\sigma\sqrt{n}Z_n +\mu n \geq \mu n+ n^a)$$ which can be rearranged as $$P(Z_n \geq \frac{n^a}{\sqrt{n}})$$ and we can simplify the expression to the right of the inequality sign as $n^{a-1/2}$. So if $a>1/2$ the right hand side will go to infinity as $n \rightarrow \infty$ so the probability goes to 0. If $a \leq 1/2$ then the right hand of the inequality sign will go to zero, so it is possible that it will have positive probability. I'm not entirely sure that you wouldn't need to put some condition on the distribution of $Z_n$ for this argument to work, but in general it seems pretty straightforward.