Central limit theorem, show that $E\left[\frac{|S_n|}{\sqrt{n}}\right]$ converges to $E[|Z|]$ This is the exercise 21.9 to the chapter "The Central Limit Theorem" in the book of Jean Jacod and Philip Protter "Probability essentials".
$\mathbf{Exercise:}$ Let $(X_j)_{j\ge1}$ be i.i.d. with $E[X_j]=0$ and $\sigma^2_{X_j}=\sigma^2<\infty$. Let $S_n=\sum_{j=1}^nX_j.$ Show that:
$\lim_{n\to\infty}E\left[\frac{|S_n|}{\sqrt{n}}\right]=\sqrt{\frac{2}{\pi}}\sigma$
$\mathbf{My}$ $\mathbf{attempt:}$ If I understand it right, we need to show that $\frac{|S_n|}{\sqrt{n}}$ converges in distribution to $|Z|$, where $Z\sim N(0,\sigma^2)$, and then it is easy to conclude that $E[|Z|]=\sqrt{\frac{2}{\pi}}\sigma$, so we would be done. Note that the central limit theorem asserts that $\frac{S_n}{\sqrt{n}}\to N(0,\sigma^2)$, which already seems to be a bit suspicious. What I've done I tried to rewrite $|S_n|$ as the sum $S_n\mathbf{1}_{S_n\ge0}+S_n\mathbf{1}_{S_n<0}$ and play with the expectation of this sum but it gave me nothing. Then I decided to work with characteristic functions but all my efforts went in vain (I got stuck in finding the char functions both of $\frac{|S_n|}{\sqrt{n}}$ and of $|Z|$). Though it is not difficult to calculate the density function of $|Z|$.
Any hint on how to proceed in solving of this exercise is desperately welcome!
 A: The part which is easy to conclude: 
Sure, if the uniform integrability is prevoiusly defined and if convergence of expectations under u.i. condition is proved then it suffices to say that uniform boundedness of second moments of $|S_n|/\sqrt{n}$ implies u.i. But as I can see, we cannot based on u.i. in frame of the book cited. Then let us prove the convergence of expectations directly.
Express the expectation of $\frac{|S_n|}{\sqrt{n}}$ in the form
$$
\mathbb E\left[\frac{|S_n|}{\sqrt{n}}\right] = \mathbb E\left[\,f_N\left(\frac{S_n}{\sqrt{n}}\right)\right] - N\mathbb P\left(\frac{|S_n|}{\sqrt{n}}\geq N\right)+\mathbb E\left[\frac{|S_n|}{\sqrt{n}}, \frac{|S_n|}{\sqrt{n}}\geq N\right] $$
$$ = E_1-E_2+E_3,
$$
where $N>0$ and continuous and bounded function $f$ is given as follows:
$$
f_N(x)=\begin{cases} |x|, & |x|<N, \cr N, & |x|\geq N\end{cases}
$$
By CLT and the definition of weak convergence, $\frac{S_n}{\sqrt{n}}\xrightarrow{d} Z\sim N(0,\sigma^2)$ implies $E_3\to \mathbb E[f_N(Z)]$ as $n\to\infty$ for any $N>0$ since $f_N$ is continuous and bounded function.
Markov inequality $\mathbb P(|X|\geq N)=\mathbb P(X^2\geq N^2)\leq \frac{\mathbb E[X^2]}{N^2}$ yields
$$
0\leq E_2=N\mathbb P\left(\frac{|S_n|}{\sqrt{n}}\geq N\right)\leq N\frac{\mathbb E\left[\frac{S_n^2}{n}\right]}{N^2}=\frac{ \text{Var}\left[S_n\right]}{nN}=\frac{\sigma^2}{N} \quad \text{for any }n\geq 1.
$$
And $E_3$ can be uniformly bounded too: 
$$
0\leq E_3=\mathbb E\left[\frac{|S_n|}{\sqrt{n}},\frac{|S_n|}{\sqrt{n}}\geq N\right]\leq \mathbb E\left[\frac{|S_n|}{\sqrt{n}}\times\underbrace{\frac{\frac{|S_n|}{\sqrt{n}}}{N}}_{\geq 1},\ \frac{|S_n|}{\sqrt{n}}\geq N\right]\leq
\frac{\mathbb E\left[\frac{S_n^2}{n}\right]}{N}=\frac{\sigma^2}{N}. 
$$
Since
$$
E_1-E_2\leq E\left[\frac{|S_n|}{\sqrt{n}}\right]\leq E_1+E_3
$$
and we bounded uniformly $E_2$ and $E_3$ for any $n$, take limits as $n\to\infty$ for $N$ fixed:
$$\tag{1}\label{1}
\mathbb E[f_N(Z)]-\frac{\sigma^2}{N}\leq\liminf_{n\to\infty} \mathbb E\left[\frac{|S_n|}{\sqrt{n}}\right]\leq\limsup_{n\to\infty} \mathbb E\left[\frac{|S_n|}{\sqrt{n}}\right]\leq \mathbb E[f_N(Z)]+\frac{\sigma^2}{N}
$$ 
Next allow $N$ tendes to infinity. Repeat the same:
$$
\mathbb E[|Z|]=\mathbb E[f_N(Z)] - N\mathbb P(|Z|\geq N)+\mathbb E\left[|Z|, \ |Z|\geq N\right].
$$
And
$$
\mathbb E[f_N(Z)] \leq \mathbb E[|Z|] + N\mathbb P(|Z|\geq N) \leq \mathbb E[|Z|] + \frac{\sigma^2}{N}
$$
and 
$$
\mathbb E[f_N(Z)]\geq \mathbb E[|Z|] - E\left[|Z|, \ |Z|\geq N\right] \geq 
\mathbb E[|Z|] -\frac{\sigma^2}{N}.$$
So, $\mathbb E[f_N(Z)]\to \mathbb E[|Z|]$ as $N\to\infty$ and limsup and liminf in (\ref{1}) tends to the same limit $\mathbb E[|Z|]$ as $N\to\infty$. 
