When does $X_n\stackrel{d}{\rightarrow}X$ and $Y_n\stackrel{d}{\rightarrow}Y$ imply $X_n+Y_n\stackrel{d}{\rightarrow}X+Y$? The question:
$X_n\stackrel{d}{\rightarrow}X$ and $Y_n\stackrel{d}{\rightarrow}Y \stackrel{?}{\implies} X_n+Y_n\stackrel{d}{\rightarrow}X+Y$
I know that this does not hold in general; Slutsky's theorem only applies when one or both of the convergences is in probability.
However, are there instances in which it does hold?
For instance, if the sequences $X_n$ and $Y_n$ are independent.
 A: Yes, independence is sufficient: The antecedent conditions here concern convergence in distribution for the marginal distributions of $\{ X_n \}$ and $\{ Y_n \}$.  The reason that the implication does not hold generally is that there is nothing in the antecedent conditions that deals with the statistical dependence between the elements of the two sequences.  If you were to impose independence of the sequences then that would be sufficient to ensure convergence in distribution of the sum.
(Alecos has added an excellent answer below that proves this result using characteristic functions.  Asymptotic independence is also sufficient for this implication, since the same limiting decomposition of the characteristic functions occurs.)
A: The Cramer-Wold theorem gives a necessary and sufficient condition:
Let $\{z_n\}$ be a sequence of $R^K$-valued random variables. Then,
$$
z_n \to_d z\;\Longleftrightarrow\;\lambda'z_n\to_d \lambda'z\quad\forall\quad \lambda\in R^K\backslash\{0\}
$$
To give an example, let $U\sim N(0,1)$ and define $W_n:=U$ as well as $V_n:=(-1)^nU$. We then trivially have $$W_n\to_d U$$ and, due to symmetry of the standard normal distribution, that
$$V_n\to_d U.$$ 
However, $W_n+V_n$ does not converge in distribution, as
$$
W_n+V_n=\begin{cases}2U\sim N(0,4)&\text{for}\;n\;\text{even}\\
0&\text{for}\;n\;\text{odd}\end{cases}
$$
This is an application of the Cramer-Wold Device for $\lambda=(1,\;1)'$.
A: Formalizing @Ben answer, independence is almost a sufficient condition, because we know that the characteristic function of the sum of two independent RV's is the product of their marginal characteristic functions. Let $$Z_n = X_n + Y_n$$. Under independence of $X_n$ and $Y_n$,
$$\phi_{Z_n}(t) = \phi_{X_n}(t)\phi_{Y_n}(t)$$
So
$$\lim \phi_{Z_n}(t) =\lim \Big [\phi_{X_n}(t)\phi_{Y_n}(t)\Big]$$
and we have (since we assume that $X_n$ and $Y_n$ converge)
$$\lim \Big [\phi_{X_n}(t)\phi_{Y_n}(t)\Big] = \lim \phi_{X_n}(t)\cdot \lim \phi_{Y_n}(t) = \phi_{X}(t)\cdot \phi_{Y}(t) $$
which is the characteristic function of $X+Y$... if $X+Y$ are independent. And they will be independent if one of the two has a continuous distribution function (see this post). This is the condition required in addition to independence of the sequences, so that independence is preserved at the limit.
Without independence we would have 
$$\phi_{Z_n}(t) \neq \phi_{X_n}(t)\phi_{Y_n}(t)$$
and no general assertion can be made about the limit.
