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Let $\{X_n\}$ be a random variable sequence and $X\sim N(0,\sigma)$. In general, the convergence $E(X_n^k) \stackrel{n}{\longrightarrow}E(X^k)$ doesn't implie that $E(X_n^{k+1}) \stackrel{n}{\longrightarrow}E(X^{k+1})$, even when we have $X_n\stackrel{D}{\longrightarrow}X$. However, there is a especific situation causing me doubts:

  1. I know that $X_n\stackrel{D}{\longrightarrow}X\sim N(0,c(1-c))$, $0<c<1$.

  2. I know that the $X_n's$ Moment generating function is $$M_{X_n}(t) = \frac{\left[1-c+c\exp\left(\frac{t}{\sqrt{n}}\right)\right]^n}{\exp(c\sqrt{n}t)}, t\in \mathbb{R}.$$

  3. It's possible realize that $$E(X_n) = \frac{d}{dt}M_{X_n}(t)|_{t=0} = 0,$$ $$E(X_n)^2 = \frac{d^2}{dt^2}M_{X_n}(t)|_{t=0} = c(1-c),$$ $$E(X_n)^3 = \frac{d^3}{dt^3}M_{X_n}(t)|_{t=0} = \frac{c^3}{\sqrt{n}}\left(2+\frac{1}{c^2}+\frac{1}{c}\right).$$ I haven't evaluated the fourth moment yet. Clearly, $E(X_n^k) \stackrel{n}{\longrightarrow}E(X^k)$ for $k=1,2,3$. I've been asking myself: In this case, can we affirm that moment's convergence for each $k$ (integer)?

Clarifications: I've been wondering about two approaches. First, to get the moment's convergence above, it's enough prove that $\sup_nE|X_n|^{k+\varepsilon}<\infty$, according to Theorem 25.12's Corollary in Billingsley's book (Probability and Measure, English version, third edition, 1995, page 338), but, until this moment, I couldn't determine the default behavior of the moment's sequence. Second, Perhaps some kind of proof using the Induction Method (about $k$) works to affirm the convergence in $(K+1)$-th order moment from the convergence of $K$-th order moment, but I need to realize the behaviour of $M_{X_n}(t)^,$s derivate. Does anyone have suggestions or ideas?

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