does asymptotic normality imply convergence of expectations? Suppose $\sqrt{n}(X_n-\mu)\to N(0,\sigma^2)$. I know this implies $X_n\to \mu$ in probability. Does it imply $E(X_n)\to \mu$? I heard about a converse to the lindebergh CLT (but couldn't find it on wikipedia), and was thinking that the lindebergh condition might be close to uniform integrability. I don't know if this works though, or if there is a simpler way.
 A: No, it does not.  It does not even imply that $E[X_n]$ exists.
[This is basically another way of writing @JohnL's answer]
First, suppose $X_n=Z_n+Y_n$ where $Z_n\sim N(0,1/n)$ and $Y_n$ is $n$ with probability $1/n$ and 0 otherwise. Note that $X_n=Z_n$ with probability $(n-1)/n$.
Let $\mu=0$. We have
$$\sqrt{n}(X_n-\mu)\stackrel{d}{\to} N(0,1)$$
Indeed,for any $t$,
$P(\sqrt{n}(X_n-\mu)<t)$ and $P(\sqrt{n}Z_n<t)$ differ by at most $1/n$,  and $\sqrt{n}Z_n\sim N(0,1)$, so the cdf of $\sqrt{n}(X_n-\mu)$  converges pointwise to the CDF of $\sqrt{n}Z_n$, which is the standard Normal CDF, and we have the required convergence in distribution (with $\sigma^2=1$)
However, $E[X_n]=E[Z_n]+E[Y_n]=0+1$ for all $n$, so $E[X_n]$ does not converge to $\mu$.
If we took $Y_n=n^2$ instead of $n$ with probability $1/n$, we would have $E[X_n]=n\to\infty$.  And if $Y_n$ was sampled from a Cauchy distribution with probability $1/n$, $E[X_n]$ would not exist for any finite $n$.
A: Let $\Phi(x)$ denote the standard normal distribution function.
Suppose $\sqrt{n} X_n$ has the distribution function:
$\Phi(x)$ if $x<n$
$\Phi(n)$ if $n \le x<\frac{\sqrt{n}}{1-\Phi(n)}$
$1$ if $x \ge \frac{\sqrt{n}}{1-\Phi(n)}$
Then, $\sqrt{n}X_n\rightarrow N(0,1)$.
Also, since $P\left[\sqrt{n}X_n=\frac{\sqrt{n}}{1-\Phi(n)} \right]=1-\Phi(n)$,
we have:
$E[\sqrt{n}X_n]=\int_{-\infty}^{n}x\phi(x)dx+\frac{\sqrt{n}}{1-\Phi(n)}P\left[\sqrt{n}X_n=\frac{\sqrt{n}}{1-\Phi(n)} \right]=-\frac{e^{-\frac{n^2}{2}}}{\sqrt{2 \pi}}+\sqrt{n}$.
Thus, $E[X_n]=\frac{1}{\sqrt{n}}E[\sqrt{n}X_n]\rightarrow 1$
