Example: convergence in distributions Give an example $X _n \rightarrow X$ in distribution, $Y _n \rightarrow Y$ in distribution, but $X_n + Y_n$ does not converge to $X+Y$ in distribution. 
I got a trivial one. $X_n$ is $\mathcal N(0,1)$ $\forall n$, $Y_n=-X_n$, $X$ and $Y$ are also $\mathcal N(0,1)$, then $X _n \rightarrow X$ and $Y _n \rightarrow Y$, but $X_n+Y_n=0$ does not converge to $X+Y$ which is $\mathcal N(0,2)$ distributed. 
Do you have a more interesting example?    

Notice that if $(X_n)$ and $(Y_n)$ are weakly convergent, then the sequence $(X_n+Y_n)_{n\geqslant 1}$ is tight, hence we can extract a weakly convergent subsequence. Therefore, the problem may come from the non-uniqueness of the potential limiting distributions. 
Consider $(\xi_i)_{i\geqslant 1}$ a sequence of i.i.d. centered random variables with unit variance.
If $n$ is even, define 
$$X_n:=\frac 1{\sqrt n}\sum_{i=1}^n\xi_i\mbox{ and }Y_n:=-\frac 1{\sqrt n}\sum_{i=1}^n\xi_i,$$
and if $n$ is odd, then 
$$X_n:=\frac 1{\sqrt n}\sum_{i=1}^n\xi_i\mbox{ and }Y_n:=\frac 1{\sqrt n}\sum_{i=n+1}^{2n}\xi_i.$$
Then $X_n\to X$ and $Y_n\to Y$ where $X$ and $Y$ are standard normal, but the sequence $(X_{2n}+Y_{2n})_{n\geqslant 1}$ is null while the sequence $(X_{2n+1}+Y_{2n+1})_{n\geqslant 1}$ converges weakly to a (non-degenerated) normal distribution.
