If a sequence of quadratic forms converges in probability and a random vector converges in distribution then $X_n^TQ_nX_n$ converges If a sequence of quadratic forms converges in probability $Q_n\xrightarrow{P}Q$ and a random vector converges in distribution  $X_n\xrightarrow{d}X$ then $X_n^TQ_nX_n\xrightarrow{d}X^TQX$.
This is a statement from an online source in statistics. It follows by Slutsky's theorem and the continuous mapping theorem. I can also see how intuitively it should be true, but I'm having trouble setting up the argument. No matter what I do, in the end I have a product of two things converging only in distribution.
 A: A first step is to prove that the sequence $\left(X_n^T(Q_n-Q)X_n\right)_{n\geqslant 1}$ converges in probability to $0$. To see this, fix a positive $\varepsilon$. There is some $R$ for which for each $n$, $\mathbb P\{\lVert X_n\rVert\gt R\} \lt \varepsilon$. Then 
\begin{align}
\mathbb P\{\left|X_n^T(Q_n-Q)X_n\right|\gt \delta\}&\leqslant 
\mathbb P\{\lVert X_n\rVert \lVert Q_n-Q\rVert \lVert X_,\rVert \gt \delta\} \\
&\leqslant 2\mathbb P\{\lVert X_n\rVert \gt R\}+\mathbb P\{\lVert Q_n-Q\rVert  \gt \varepsilon/R^2\}\\
&\leqslant 2\varepsilon+ \mathbb P\{\lVert Q_n-Q\rVert  \gt \varepsilon/R^2\}.
\end{align}
Therefore, the question reduces to the case where $Q_n=Q$ for each $n$.
It is true that the sequence $\left(X_n^TQX_n\right)_{n\geqslant 1}$ is tight, hence it admits a subsequence which converges in distribution.
But if $X$ is a symmetric non-degenerated random variable and $X_n:=e^{(-1)^nX}$, then $X_n$ has the same distribution as $e^{X}$; if $Q=e^{2X}$, then the distribution on $X_nQX_n$ is that of $1$ or $e^{4X}$, hence this does not converge in distribution. 
However, we do have $X_n^TQX_n\to X^TQX$ if $Q$ is not random, since we can use the continuous mapping theorem with $x\mapsto x^TQx$.
A: Write $Q_n = \Sigma_n^T\Sigma_n \xrightarrow{p}\Sigma^T\Sigma = Q$
By Slutsky theorem $X_n^T\Sigma_n^T \xrightarrow{d} X^T\Sigma^T = (\Sigma X)^T$
Similarly, $\Sigma_n X_n\xrightarrow{d} \Sigma X$.
Now $(\Sigma X)^T\Sigma X$ is a quadratic polynomial in $\dim(X)$ variables, but each of its terms is a monomial in one variable. So the continue mapping theorem applies.
