TITLE: Convergence of moments implies convergence to normal distribution
QUESTION [12 upvotes]: I have a sequence $\{X_n\}$ of random variables supported on the real line, as well as a normally distributed random variable $X$ (whose mean and variance are known but irrelevant). I know that the moments of the $X_n$ converge to the corresponding moment of $X$, that is, for every $k\ge1$,
$$
\lim_{n\to\infty} \mu_k'(X_n) = \mu_k(X).
$$
I need to conclude that the $X_n$ converge to $X$ in distribution.
I believe that this is a standard fact in probability, and I would like an excellent source (including a clear statement and proof) for this fact, to cite in a paper I'm writing. (The application is to number theory, which is why I added the probabilistic-number-theory tag.) I also believe that this conclusion holds for many, but not all, random variables $X$ and not just a normally distributed one; I'd be happy for a general statement or one that applies only to a normal variable.
Nominations for a good citing source, anyone?

REPLY [11 votes]: It is theorem 30.2 in Billingsley's Probability and Measure (I own a second Polish edition, so numbering may differ a little).
It's quite easy to prove it, once you estabilish Prokhorov's theorem; namely use boundedness of some moments to conclude that your sequence of distributions is tight and then it suffices to convey everyone that every convergent subsequence of $(X_n)$ converges to $X$ (because convergence in distribution is metrizable), which is easy, because the limit is characterized by its moments. Before that, one needs a lemma stating that convergence in distribution combined with convergence of moments implies that moments converge to the moments of the limit.