Show $n^{-p}*S_{n} \to 0$ in probability Suppose we have a sequence of i.i.d random variables $\{X_{n}\}_{n\in \Bbb N}$ with mean $0$ and variance $\sigma^{2} < \infty$. We want to show that $n^{-p}S_{n} \to 0$ in probability when $p>\frac{1}{2}$.
Here's my attempt

Showing $n^{-p}S_{n} \to 0$ in probability is equivalent to showing $$\Bbb P\left(\left|\frac{S_{n}}{n^{p}}\right|\leq \epsilon\right) \to 1 $$
Rearrange the expression in the bracket we finally get
  $$\Bbb P\left(\left|\frac{S_{n}}{\sigma \sqrt{n}}\right| \leq \frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right).$$
By central limit theorem, we know $\frac{S_{n}}{\sigma \sqrt{n}} \to N(0,1)$ in distribution. But my question is that can we directly write the expression above as $$2\Phi \left(\frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right)-1$$ and argue that as $n \to \infty$, we get $$\Bbb P\left(\left|\frac{S_{n}}{\sigma \sqrt{n}}\right| \leq \frac{\epsilon}{\sigma}n^{p-\frac{1}{2}}\right) \to 1$$
I don't know if central limit theorem is still valid when the expression in the bracket changes, Could someone help me with it? if not, how to prove it

 A: The central limit theorem gives that for all fixed point $t$, 
$$\tag{*}
\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant t\right)\to 
\mathbb P(\lvert N\vert\leqslant t)
$$
where $N$ has a standard normal distribution. 
Actually, one can show that the convergence is also uniform on $\mathbb R$ because the limiting distribution function is continuous.
We can also conclude without using the uniform convergence: fix $R\gt 0$; for $n$ 
large enough, $\epsilon n^{p-1/2}\geqslant R$ hence 
$$
\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant R \right).
$$
Taking the $\liminf_{n\to+\infty}$ and using (*) with $t=R$ gives 
$$
\liminf_{n\to+\infty}\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant\mathbb P\left(\left\lvert N\right\rvert\leqslant R \right).
$$
Now, since $R$ the previous inequality is valid for all $R$, we derive that 
$$
\liminf_{n\to+\infty}\mathbb P\left(\left\lvert \frac{S_n}{\sigma\sqrt n}\right\rvert\leqslant \epsilon n^{p-1/2} \right)\geqslant1.$$
A: We can write $\frac{1}{n^p}S_n=\frac{1}{n^{p-1/2}}\frac{{S_n}}{\sqrt{n}}$
$\frac{{S_n}}{\sqrt{n}}$ converges in distribution to $N(0,\sigma^2),$ and $\frac{1}{n^{p-1/2}}$ converges in distribution to $0$ then $\frac{1}{n^p}S_n$ converges in distribution to $0$ which means convergence in probability to $0$. 
