if $\{x_i\}$ is a sequence of zero mean independent random variables that satisfies the Lyapunov condition:$$s_n^{-2r}\sum_{i=1}^nEx_i^{2r} \to 0$$ as $n \to \infty$. here $s_n^2=\sum_{i=1}^nEx_i^2$, r is some integer greater than 1. let $S_n=\sum_{i=1}^nx_i$, how to prove that:$$\lim_{n \to \infty}E[(\frac {S_n}{s_n})^{2r}] = E(Z^{2r})$$ where $Z\sim N(0,1)$.
Converge in distribution doesn't imply converge of mean, but Lyapunov conditionv is stronger than converge in distribution.