$\{X_n\}$ i.i.d. with mean zero and finite variance. Prove that $\lim\limits_n E[|S_n|/\sqrt n]=2\lim\limits_n E[S_n^+/\sqrt n]=\sqrt{2/\pi}\sigma$ Let $\{X_n\}$ i.i.d. with mean zero and variance $\sigma^2$,$0<\sigma^2<\infty$.$S_n=\sum_{i=1}^nX_i$ Prove that $\lim\limits_n E[|S_n|/\sqrt n]=2\lim\limits_n E[S_n^+/\sqrt n]=\sqrt{2/\pi}\sigma$
From CLT we can get that $S_n/\sigma\sqrt n$ converges in distribution to $\mathcal{N}(0,1)$,then does $S_n^+/\sigma\sqrt n$ converge in distribution to $\mathcal{N}^+(0,1)$? Then suppose that statement holds true, is it correct to say $\lim\limits_n E[S_n^+/\sigma\sqrt n]=E[\mathcal{N}^+(0,1)]$?
 A: In general, if $M_n\stackrel{d}{\Longrightarrow} M$, $\mathbb{E}[|M|]\leq\liminf\mathbb{E}[|M_n|]$. If $M_n$ is uniformly integrable, then equality holds with $\lim_n$ instead of $\liminf_n$(see this posting and references).
Let $M_n=\frac{X_1+\ldots+X_n}{\sqrt{n}}=\frac{S_n}{\sqrt{n}}$. Notice that for any $n$,
$$\mathbb{E}[|M_n|^2]=\sigma^2$$
whence we obtain
\begin{align}
\mathbb{E}\Big[|M_n|\mathbb{1}(|M_n|>a)\Big]&\leq\frac{1}{a}\mathbb{E}[|M_n|^2]=\frac{\sigma^2}{a}
\end{align}
Consequently
$$\lim_{a\rightarrow\infty}\sup_n\mathbb{E}\Big[|M_n|\mathbb{1}(|M_n|>a)\Big]=0$$
Therefore
$$\lim_n\mathbb{E}\Big[\big|\frac{S_n}{\sqrt{n}}\big|\Big]=\frac{\sigma}{\sqrt{2\pi}} \int|x|e^{-x^2/2}\,dx=\sigma\sqrt{\frac{2}{\pi}}$$
A similar argument applies to $\frac{S^+_n}{\sqrt{n}}=f(M_n/\sqrt{\sigma})$ where $f(x)=x_+$, since $\mathbb{E}[M^+_n|^2]\leq \mathbb{E}[|M_n|^2]$.
The CLT yields the $M^+_n=f(M_n)\stackrel{d}{\Longrightarrow}f\big(N(0,\sigma^2)\big)$. The uniformly integrable condition is obvious, (here $f(N(0,\sigma^2))$ denotes the law of a random viable $Y=X_+$ where $X\sim N(0,1)$, that is the half-normal distribution.)
