If $Z\_1,Z\_2,Z\_3,\ldots$ are i.i.d. with $P(Z\_i=-1) = P(Z\_i=+1) = \frac 12,$ then we have by the Central Limit Theorem that $\frac{\sum\_{i=1}^n Z\_i}{\sqrt{n}}\stackrel{d}{\to} \mathcal{N}(0,1),$ so that for any continuous bounded function $f,$ we have $\mathbb{E}f\left(\frac{\sum\_{i=1}^n Z\_i}{\sqrt{n}}\right)\to\mathbb{E}f(W)$ where $W\sim\mathcal{N}(0,1).$ Now, $|\cdot|$ is not a bounded function, so it is not necessarily true that


$$\mathbb{E}\left|\frac{\sum\_{i=1}^n Z\_i}{\sqrt{n}}\right|\to\mathbb{E}|W|.$$


My question is whether the above is true for this specific distribution of $Z\_i.$ If not, what does $\mathbb{E}\left|\frac{\sum\_{i=1}^n Z\_i}{\sqrt{n}}\right|$ converge to?



Central limit theorem and convergence of means
The following result appears in Billingsley's book *Convergence of probability measures* (1968, and I guess the 1995 edition).



> 
> **Theorem.** Let $(Y\_n)\_{n\geqslant 1}$ be a sequence of non-negative random variables defined on a probability space $(\Omega,\mathcal F,\mu)$ such that $Y\_n\to Y$ in distribution and $Y\_n,Y$ have a finite expectation. Then the following are equivalent:
> 
> 
> i) $\mathbb E[Y\_n]\to\mathbb E[Y]$;
> 
> 
> ii) the sequence $(Y\_n)\_{n\geqslant 1}$ is uniformly integrable.
> 
> 
> 


Assume that i) holds and fix a positive $\varepsilon$. Fix $R$ such that $\mathbb E[Y\chi\_{\{Y>R\}}]\lt\varepsilon$ and $\mu\{|Y|=R\}=0$. The set of discontinuity points of map $t\mapsto \chi\_{-(R,R)}(t)$ has a null $\mu\_Y$ measure, hence
$$\lim\_{n\to\infty}\mathbb E[Y\_n\chi\_{\{Y\_n\lt R\}}]=\mathbb E[Y\chi\_{\{Y\lt R\}}].$$


Conversely, we use the map $t\mapsto \chi\_{-(R,R)}(t)$ for a well-chosen $R$.


In the context of the question, with $Y\_n:=n^{-1/2}\left|\sum\_{j=1}^nX\_j\right|$, we observe that $\mathbb E[Y\_n^2]=\mathbb E[X\_0^2]$ (it holds more generally in the case of an iid sequence of zero-mean square integrable functions).


