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 $E\left\{(X\_{n})^{K}\right\} \stackrel{n}{\longrightarrow} E(X^{K})$, where $K\geq 1$ is an integer number. Can we say that $E\left\{(X\_{n})^{K+1}\right\} \stackrel{n}{\longrightarrow} E(X^{K+1})?$


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



convergence in distribution and convergence of moments
No, this is not true. Let $X \sim N(0,1)$ and define $Y\_n$ to be independent of $X$, such that $Y\_n = \sqrt{n}$ with probability $1/n$ and 0 otherwise. Set $X\_n = X+ Y\_n$. Since $Y\_n \to 0$ in $L^1$, we have $X\_n \to X$ in $L^1$ and hence also in distribution; in particular, $E X\_n^1 \to E X^1 = 0$. But by independence $E[X\_n^2] = E[X^2] + E[Y\_n^2] = 1+1 = 2$ for all $n$, whereas $EX^2 = 1$.


