Convergence in distribution plus convergence of moments.

Question

Suppose that the sequence of r.v $\{X_n\}_{n\geq 1}$ has all the moments, and $X_n\stackrel{D}{\longrightarrow} X\sim N(0,\sigma)$. Assume that $E\{(X_n)^K\} \stackrel{n} {\longrightarrow} E(X^K)$, where $K\geq 1$ is an integer number. Can we say that $E\{(X_n)^{K+1}\} \stackrel{n}{\longrightarrow} E(X^{K+1})?$

Clarifications: The simbol $\stackrel{D}{\longrightarrow}$ represents convergence in Distribution.

Answer 1

It is not true in general: take $X_n=X+Y_n$, where $\mathbb E[|Y_n|^K]\to 0$ but $\mathbb E[|Y_n|^{K +1}]=1$ for each $n$.