Given the random variables $(X_n)_{n\in \Bbb N^*}$ such that $\sqrt{n}X_n$ converges in distribution to the standard normal distribution $$\sqrt{n}X_n \xrightarrow[n \to +\infty]{\mathcal{D}} \mathcal{N}(0,1)$$
Suppose $E(|X_n|) <+\infty$ for all $n\in \Bbb N^*$.
Prove that
$$\sup_{n\in \Bbb N^*}E(|X_n|)<+\infty$$
In other words, there exist an $M <+\infty$ such that $E(|X_n|)<M$ for all $n\in \Bbb N^*$.
My attempt:
Let's write $$E(|X_n|) = E(|X_n|\mathbb{I}_{\{|X_n|\le M_1\}})+ E(|X_n|\mathbb{I}_{\{|X_n|>M_1\}}) $$ with $M_1>0$
The first term is bounded by $M_1$.
About the second term, from an $N$ sufficiently large, we have intuitively $X_n \sim\frac{1} {\sqrt{n}}\mathcal{N}(0,1) $ then $f_{X_n}(x) ~\approx \frac{\sqrt{n}}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2n} \right)$ hence the linear increasing of $x$ is much less than the decreasing of $f_{X_n}(x)$ for $x>M_1$. Hence, $E(|X_n|\mathbb{I}_{\{|X_n|\le M_1\}})$ will be bounded by a constant $M_2$ for all $n> N$.
But I haven't been able to formalize this intuition.
Let $X\sim N(0,1)$. Let $Y_n$ take the values $0$ and $n^{2}$ with probabilities $1-\frac 1 {\sqrt n}$ and $\frac 1 {\sqrt n}$ respectively. Let $X_n=\frac X {\sqrt n}+\frac {Y_n} {\sqrt n}$. Since $P(|Y_n| >\epsilon)=\frac 1 {\sqrt n} \to 0$ it follows that $\sqrt n X_n \to X$ in distribution. Howvever, $EX_n \geq 0+\frac {n^{2}} {\sqrt n}\frac 1 {\sqrt n} \to \infty$.