I have encountered the following problem, could someone provide me some hints on how to solve it?
Assume that the sequence $(X_n)$ is i.i.d. with mean $0$ and variance $1$. For every $n\ge1$, let $S_n=X_1+\cdots+X_n$. Then, $$\lim_{n\rightarrow\infty }\frac{1}{\sqrt{n}}\ E|S_n|=\sqrt{\frac{2}{\pi}}.$$
Define $Y_n:=n^{-1/2} \left|S_n \right|$ and $Z$ a random variable of standard normal distribution. Then for a fixed $R$, $$ \left|\mathbb E(Y_n)-\mathbb E(Z)\right| \leqslant \int_0^R \left|\mathbb P\{Y_n \gt t\}-\mathbb P\{Z \gt t\}\right|\mathrm dt+ \int_R^{+\infty}\mathbb P\{Y_n \gt t\}\mathrm dt+\int_R^{+\infty}\mathbb P\{Z \gt t\}\mathrm dt.$$ Since $ \mathbb E [Y_n^2 ]=1$, we get by Cauchy-Schwarz inequality $$\int_R^{+\infty}\mathbb P\{Y_n \gt t\}\mathrm dt=\int_R^{+\infty}\mathbb P\{Y_n \mathbf 1 \{Y_n \gt R\} \gt t\}\mathrm dt \leqslant \mathbb E\left[Y_n\mathbf 1 \{Y_n \gt R\}\right] \leqslant \mu\{Y_n \gt R\}^{1/2}$$ we obtain by Markov's inequality $$\int_R^{+\infty}\mathbb P\{Y_n \gt t\}\mathrm dt \leqslant \frac 1R,$$ and a similar estimate holds for $Z$. In view of the first displayed equation, it follows that $$ \tag{*}\left|\mathbb E(Y_n)-\mathbb E(Z)\right| \leqslant R \sup_{t\in [-R,R ] }\left|\mathbb P\{Y_n \gt t\}-\mathbb P\{Z \gt t\}\right|+ \frac 2R.$$ Since the distribution function of $Z$ is continuous, the first term of (*) converges to $0$ as $n$ goes to infinity.