$\mathbb P $- convergence implies $L^2$-convergence for gaussian sequences Consider $(X_n)_{n \in \mathbb N}$ a sequence of gaussian random variables whose limit in probability exists and is given by $X$.
I was interested in showing that in this particular case we have always that this sequence converge also in $\mathbb L ^2$ ($\mathbb L ^p$) towards $X$.
Is the following arguments sufficient to show such fact (since $X$ is gaussian as a limit in probability of $X_n$ and $(X_n), X \in \mathbb L^p$)? 
Given an arbitrary $\epsilon >0$
\begin{align} \mathbb E [ (X_n -X)^2]&= \mathbb E [ (X_n -X)^2\mathbf 1_{|X_n -X|> \epsilon}]+\mathbb E [ (X_n -X)^2\mathbf 1_{|X_n -X|< \epsilon}]\\& \leq\mathbb E [ (X_n -X)^4]^{1/2}\mathbb P(|X_n-X|> \epsilon)+ \epsilon^2
\end{align}
Also, I have a vague memory of a minimal condition to this implication in the general case when de sequence is not necessarily normal distributed. Should it be that the sequence is almost surely bounded ?
Many tank's for your help.

We have to show that for any subsequence $(X_{n_k})$ we can extract a further subsequence $(X_{n'_k})$ such $\mathbb E|X_{n'_k}-X|^p\to 0$. We can assume that $X_{n_k}\to X$ almost surely by passing to a further subsequence. Then $X$ is Gaussian. 
Write $X_n=a_nN+b_n$, $X=aN+b$, where $N\sim N(0,1)$. 
If $a\gt 0$, then $a_n\gt a/2$ for $n$ large enough hence we can prove that $\sup_n\mathbb E|X_n|^p$ is finite for each $p$. We then conclude as in the OP attempt. 
