Let $X_n$ be a sequence of random variables such that $$\frac{X_n}{s\log^2 n}\rightarrow X$$ converges in $L^p$ for all $p>0$ to a random variable $X$ for an $s\in(0,\infty)$ and $\sigma^2_X\in(0,\infty)$ the variance of $X$.
How to show that $\mathbb{E}(|X_n-\mathbb{E}(X_n)|^2)\sim (s\sigma_X)^2\log^4 n$?
So we want to show that $$\lim\limits_{n\rightarrow\infty}\frac{\mathbb{E}(|X_n-\mathbb{E}(X_n)|^2)}{(s\sigma_X)^2\log^4 n}=1$$ using $$\lim\limits_{n\rightarrow\infty}\mathbb{E}\left(|\frac{X_n}{s\log^2 n}-X|^p\right)=0.$$ How do I do this?
Observe that $$\mathbb{E}(|X_n-\mathbb{E}(X_n)|^2)=\mathbb{E}\left[X_n^2\right ]-\left( \mathbb{E}\left[X_n\right ]\right) ^2 $$ and using the convergences in $\mathbb L^1$ and $\mathbb L^2$, one gets sequences $\left(\delta_n \right)_{n\geqslant 1} $ and $\left(\varepsilon_n \right)_{n\geqslant 1} $ converging to zero such that $$ \mathbb{E}\left[X_n^2\right ]=\mathbb E\left[X^2\right] s^2\log^4n\left(1+\delta_n \right) \mbox{ and } $$ $$ \mathbb{E}\left[X_n\right ]=\mathbb E\left[X\right] s\log^2n\left(1+\varepsilon_n \right). $$ Consequently $$\mathbb{E}(|X_n-\mathbb{E}(X_n)|^2)= s^2\log^4n\left(\mathbb{E}\left[X_n^2\right ]-(\mathbb E\left[X\right])^2+\delta_n\mathbb E\left[X^2\right]+ (2\varepsilon_n+\varepsilon_n^2 ) (\mathbb E\left[X\right])^2\right) =s^2\log^4n\left(\sigma^2_X+a_n\right) $$ where $a_n \to 0$ as $n$ goes to infinity.