Show that $\lim_{n \rightarrow \infty} \frac{S_n}{n^p} = 0$ in probability

Question

i have: $(X_n)$, a sequence of iid rv with $E(X_i) = 0$ and $Var(X_i) = C$ with $C < \infty$. $p > 1/2$ and $S_n = X_1 + ... + X_n$. I want to show that $\lim_{n \rightarrow \infty} \frac{S_n}{n^p} = 0$ in probability. I tried to use the Central Limit Theorem but I don't exactly know how.

Answer 1

First, compute the variance of $\frac{S_n}{n^p}$: $$Var[\frac{S_n}{n^p}] = \frac{n Var(X_i)}{n^{2p}} = n^{1-2p}C \to 0$$ as $n\to \infty$, since $p>\frac 1 2 $.

Therefore $\frac{S_n}{n^p} \to 0$ in probability (because $L_2$-convergence implies convergence in probability).