Counter-examples related to Slutsky's Theorem. 
Consider some  real-valued random variables $(X_n)$, $(Y_n)$, $X$ and $Y$, defined on the same probability space. What is a counterexample to the claim that $X_n\to X$ in distribution and 
  $Y_n \to Y$ in distribution implies that $X_n + Y_n \to X + Y$ in distribution? 

I know that Slutsky's theorem guarantees the implication when $Y = c$ holds, but not otherwise.
 A: If the pair $(X_n,Y_n)$ converges in distribution to $(X,Y)$, then necessarily $X_n+Y_n$ will converge in distribution to $X+Y$ (by the continuous mapping theorem). So a counterexample would require us to specify $X$ and $Y$ and their joint  distribution such that $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$.
So take $X$ to be a nonconstant symmetric random variable, define $X_n:=X$, $Y_n:= X$, and $Y:=-X$. Then trivially $X_n$ converges in distribution to $X$, and $Y_n$ converges in distribution to $Y$ (since $X$ is symmetric). But $X_n+Y_n$ equals $2X$, while $X+Y=0$. Note that, as expected, $(X_n,Y_n)$ does not converge in distribution to $(X,Y)$, which is concentrated on the line $y=-x$; it converges in distribution to $(X,X)$, which is concentrated on the line $y=x$.
A: Let $Z_k$ be iid with mean $0$ and variance $1$, and let $X_n = \frac{1}{\sqrt n} \sum_{k=1}^n Z_k$ and $Y_n = -X_n$. By the CLT, both $X_n$ and $Y_n$ converge in distribution to some $X$ which is standard normal. But $X_n + Y_n$ is always zero.
A: I think the essence here is not convergence of sequences, but rather the difference between equality in distribution (denoted by $\sim$ in what follows) and almost sure equality (denoted by $=$ in what follows).
The following example, similar to what @grand_chat wrote down, should make it clear:
Let $Z \sim N(0, 1)$. Furthermore let $X = Y = A = Z$ and $B = -Z$.
Then since $Z \sim -Z$ we must have that $X \sim A$ and $Y \sim B$, but it does not hold that $X + Y \sim A + B$ since $X + Y \sim 2 \cdot N(0, 1)$ and $A + B = 0$.
