The Product of Two Sequences Converging in Distribution? Suppose there are two i.i.d. sequences $\left\{X_n\right\}$ and $\left\{Y_t\right\}$ that
\begin{align*}
\sqrt{N}\left(\bar{X}_N-\mathrm{E}\left[X_n\right]\right)\overset{d}{\rightarrow}&\mathcal{N}\left(0,\mathrm{Var}\left[X_n\right]\right)\\
\sqrt{T}\bar{Y}_T\overset{d}{\rightarrow}&\mathcal{N}\left(0,\mathrm{E}\left[Y_t^2\right]\right)
\end{align*}
and they are independent. Can I infer anything about
\begin{align*}
\frac{1}{\sqrt{NT}}\sum_{i=1}^{N}{\sum_{t=1}^{T}{X_nY_t}},
\end{align*}
i.e. the product of two sequences that converge in distribution? I think I cannot apply Slutsky's Theorem since both converge not in probability, but in distribution.  
Here I am assuming the existence of the moments $\mathrm{E}\left[X_n\right]$, $\mathrm{E}\left[X_n^2\right]$, $\mathrm{E}\left[Y_t\right]$ and $\mathrm{E}\left[Y_t^2\right]$, but, if it is necessary, then I can add the assumption that both $X$ and $Y$ are normally distributed.  
By the way, I tried Monte Carlo Simulation and it seems the product follows a leptokurtic distribution that is similar to Laplace Distribution.
 A: Neglecting centerings and so on, if $U_m$ converges in distribution to $U$, and if $V_n$ converges in distribution to $V$, and it the $U$s are independent of the $V$s, then I think it's true that the vectors $(U_m,V_n)$ converge in distribution to $(U,V)$ as $(m,n)\to(\infty,\infty)$, and hence $U_nV_n$ converges in distribution to $UV$.
One can, for instance, pretend  that the $U_m\to U$ almost surely (on some other probability space, for which all the marginal distributions are unchanged, say), and similarly for $V_n\to V$ a.s., and so $U_mV_n\to UV$ a.s. and hence in distribution.
Taking centerings and scalings into account:  Let $\sigma^2 = \mathrm{Var}(X_1)$, let $\tau^2=\mathrm{Var}(Y_1)$.
Write $\overline X_N = \mu + \sigma Z_N/\sqrt N$ and $\overline Y_T = \tau W_T/\sqrt{T}$ where $(Z_N,W_T)$ converges in distribution to a bivariate Gaussian with unit covariance matrix and zero mean. Then $S=\sqrt{NT}\overline X_N \overline Y_T $, which the original question is about, equals $\sqrt N \mu \tau W_T + \sigma\tau Z_N W_T$.  If $\mu\ne0$, $S/\sqrt N$ is asymptotically $N(0,\mu^2\tau^2)$.  If $\mu=0$, $S$ is asymptotically distributed as the product of an $N(0,\sigma^2)$ times an independent $N(0,\tau^2)$.  Note that the correct scaling on $S$ depends on whether $\mu=0$ or not.  Does this agree with your Monte Carlo experiments?
