$E\big|\overline{X}_n-\mu\big|=O(n^{-1/2})$

Question

Let $X_1,X_2,...$ be iid random variables with mean $\mu$ and variance $\sigma^2$. I'm trying to show that $$E\big|\overline{X}_n-\mu\big|=O(n^{-1/2}).$$ That is equivalent to show that for $n$ sufficiently large $$E\big|\overline{X}_n-\mu\big|\leq c\dfrac{1}{\sqrt{n}},$$ for some constant $c$. The only idea that came to my mind for proving it is the Central Limit Theorem. I know that $\sqrt{n}(\overline{X}_n-\mu)/\sigma\overset{d}{\rightarrow}N(0,1)$. Does that implies that for $n$ sufficiently large $$E\big|\overline{X}_n-\mu\big| =E\big|Z\big|\sigma\dfrac{1}{\sqrt{n}}\quad ?$$

Answer 1

By Jensen's inequality $$ E|\bar X_n - \mu| \le \sqrt{E|\bar X_n - \mu|^2} = \frac{\sigma}{\sqrt n}. $$ Hence $E|\bar X_n - \mu| = O(n^{-1/2})$ with $c = \sigma$.

Answer 2

If $Y_n \to Y$ in distribution and $EY_n^{2}$ is bounded then $E|Y_n-Y| \to 0$. We can apply this to conclude that $\sqrt n (\overline {X_n} -\mu) /\sigma \to N(0,1)$ in $L^{1}$.