When weak convergence implies moment convergence? Given a sequence $(\mu_n)_n$ of probability measures on $\mathbb R$, which converges weakly to a probability measure $\mu$, when do we have
$$
\tag{1} \lim_{n}\int x^kd\mu_n(x)=\int x^k d\mu(x) \qquad \forall k\geq 0\;?
$$
Is "$\mu$ has compact support" a sufficient condition?
Note that $\mu_n$ converges to $\mu$ weakly if
$$ \int \varphi d\mu_n \to \int \varphi d\mu$$
for all $\varphi$ which is continuous and has compact support. Note that $x^k$ are continuous but not of compact support, so (1) is not immediately obvious.
 A: This is almost 10 years late. Possibly by know you are Student no longer. In any event, if $\mu_n\stackrel{n}{\Longrightarrow\infty}\mu$, a sufficient condition for $E_{\mu_n}[X]\rightarrow E_\mu[X]$ is that
$$\lim_{a\rightarrow\infty}\sup_n\int_{\{|x|>a\}}|x|\,\mu_n(dx)=0$$
This is equivalent to say that  a sequence of random variable $X_n\sim \mu_n$, defined on a common domain $(\Omega,\mathscr{F},\mathbb{P})$ is uniformly integrable. This can be seen for instance in Billingsley, P. Convergence in Probability Measures, John Wiley & Sons, 1968, pp. 32.
From this result, is not difficult to derived sufficient conditions for $E_{\mu_n}[X^r]\xrightarrow{n\rightarrow\infty}E_{\mu}[X^r]$ (again, uniform integrability is the key).
In the counterexample given by the responder (user940), $\mu_n=(1-\tfrac1n)\delta_0+\frac1n\delta_{x_n}$ with $x_n\xrightarrow{n\rightarrow\infty}\infty$
Notice that given $a>0$,  for  all $n$ large enough $x_n>a$ and so, $$\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)=\frac{1}{n}x_n$$
If $x_n\geq n$ for all $n$, then
$$\sup_n\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)\geq1$$
This means that in this case, the measures do not satisfy the uniformintegrability condition.
A: A condition on the limit measure will never be enough.
The sequence $\left(1-{1\over n}\right)\delta_0+{1\over n}\delta_{x(n)}$ converges to $\delta_0$ weakly, but we can make its moments behave horribly by choosing $x(n)$ to be 
very large.  
A sufficient condition for your moments to converge is if all the $\mu_n$s have the same compact support.
