# 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 $$\lim_{n}\int x^kd\mu_n(x)=\int x^k d\mu(x) \qquad \forall k\geq 0\;?$$ Is "$\mu$ has compact support" a sufficient condition ?

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I'm sure you might have guessed ... But I'm editing it, thx. – Student Jun 13 '12 at 13:09

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.