Central limit theorem

From the Lindeberg-Levy central limit theorem, if $\{X_1,\ldots,X_n\}$ is a sequence of independent and identically distributed random variables with $EX_i = \mu$ and $\mathrm{Var}X_i = \sigma^2 <\infty,$ then as $n\to\infty$ the random variables converge in distribution as follows $$\frac{X_1+\ldots +X_n - n\mu}{\sqrt{n}} \to N(0,\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{x^2}{\sigma^2}}$$

I have two questions about the results of this theorem.

(1) From here what can be said about the convergence of $$\Bigg( \frac{X_1+\ldots +X_n - \mu}{\sqrt{n}}\Bigg)^k$$ where $k\in\mathbb{N}$.

Is it as simple as raising $\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{x^2}{\sigma^2}}$ to the k-th power? So as $n\to\infty$, $$\Bigg( \frac{X_1+\ldots +X_n - \mu}{\sqrt{n}}\Bigg)^k \longrightarrow \Big(\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{x^2}{\sigma^2}}\Big)^k?$$

(2) What can we say about $$E \Bigg( \frac{X_1+\ldots +X_n - n\mu}{\sqrt{n}}\Bigg)^k$$ as $n\to \infty?$ Does the argument still converge to the normal distribution or a variation of the normal distribution as $n\to\infty$?

• (1) en.wikipedia.org/wiki/Continuous_mapping_theorem (2) The expectation is identically 0 (use linearity of expectation). – Batman Nov 5 '16 at 0:33
• " Is it as simple as raising $\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{x^2}{\sigma^2}}$ to the $k-$th power " ? Why would it? That's not even a valid distribution. – leonbloy Nov 5 '16 at 0:34
• BTW, you are missing a $n$ ( $\cdots - n\mu$) in the first fraction. – leonbloy Nov 5 '16 at 0:35
• Your second question is rather trivial, and suggests that you have not yet grasped what the CLT says. Call $Z= (X_1+ \cdots + X_n - n \mu)/\sqrt{n}$. THen, if you know that $Z$ tends in distribution to a normal of mean 0, then trivially you know that $E(Z) \to 0$. Further, this later result (apart from being contained in the CLT) is much simpler, almost direct. – leonbloy Nov 5 '16 at 0:38
• @leonbloy I should rewrite my second question because I'm actually interested in knowing $E(Z)^k$ where $Z$ is the same one from your comment. – lvxvl Nov 5 '16 at 0:46

Letting $$Z= \frac{X_1+ \cdots + X_n - n \mu}{\sqrt{n}}$$ and knowing that $Z \to N(0, \sigma^2)$ then we know that $E(Z^k)\to \sigma^k (k-1)!!$ if $k$ is even, $0$ otherwise.