Convergence of expectation Suppose we have $X_n \overset{D}\to X$ for some sequence $X_{1},\dotsc, X_{n}$. Is it the case that if $E(X_{n}^2) \to E(X^2)$ we have it that $E(X_n) \to E(X)$, and when would it hold?
My first thought is that since we have $E(X_{n}^2) \to E(X^2)$ and convergence in distribution, we might somehow have convergence in mean square which would imply we have convergence in mean.
 A: We may have a sequence $(X_n)_{n\geqslant 1}$ which converges in probability to $0$ but such that $\lim_{n\to +\infty}\mathbb E\left[X_n^2\right]$ is not $0$. For example, consider the set of positive integers and the probability measure $\mathbb P\{k\}=2^{-k}$, $k\geqslant 1$. Then define the random variable 
$X_n:=c_n\mathbb 1\{j\geqslant n\}$, where $\mathbf 1$ denotes the indicator function. For any choice of $c_n$, the sequence $(X_n)$ converges in probability to $0$. But given any positive $\lambda$, we can choose the sequence $(c_n)$ such that $\mathbb E\left[X_n^2\right]=\lambda$. 
With other choices of $(c_n)$, the sequence $\left(\mathbb E\left[X_n^2\right]\right)$ may even be unbounded, or oscillated between two or more values. 
However, if $\lim_{R\to +\infty}\sup_{n\geqslant 1}\mathbb E\left[X_n^2\mathbf 1\{X_n^2\geqslant R \}\right]$ (uniform integrability), then we do have $\mathbb E\left[X_n^2\right]\to \mathbb E\left[X^2\right]$.
A: Using Skorokhod's representation theorem, we may assume $X_n \to X$ almost surely without loss of generality. 
Note also that $\vert X_n \vert \leq 1 + X_n^2$ and that, by assumption, $E(1 + X_n^2) \to 1 + EX^2$. Thus, assuming $EX_n^2 < \infty$, we get $EX_n \to EX$ using a dominated convergence theorem.
Note that this does not contradict the point made in other answers, namely that convergence in distribution alone, or in probability for that matter, does not imply convergence of $EX_n$ to $EX$.
A: In addition to the other excellent answer, I will give a simple illustration explaining why you cannot expect (always) the means to converge, even if you have convergence in distribution.  There must be some extra conditions.  Look at the figure below:

If we simulate the angle $\theta$ uniformly in $(0,\pi)$ then the point with $x$ coordinate $x=\frac1{\tan\theta}$ on the upper horizontal line will have the Cauchy distribution. 
Define now a sequence of random variables with $X_n=\min(\max(-n, \frac1{\tan\theta}),n)$ (capping at $-n,n$) (for $n=1,2,\dotsc$).  Then $X_n$ will converge in distribution to the Cauchy distribution, but $\DeclareMathOperator{\E}{\mathbb{E}} \E X_n$ will converge to zero, since by symmetry $\E X_n =0$ for all $n$ (all limits when $n \rightarrow \infty$).   But the expectation of the Cauchy distribution do not exist.  
