Is the existence of a finite variance enough for a CLT?

Question

Suppose $X_1,X_2,...$ are random variables (not neccessarily i.i.d.) with $\mathbb{E}[X_i]=0$ such that $\operatorname{Var}\left(\frac{X_1+...+X_n}{\sqrt{n}}\right)\underset{n \longrightarrow \infty}{\longrightarrow}C$, $C>0$.

Is this enough to deduce a CLT?

Answer 1

Let $Y$ be any random variable with mean $0$ and finite variance, $X_1=Y$ and $X_n=(\sqrt n -\sqrt {n-1})Y$, for $n \geq 2$. Then $\frac {X_1+X_2+\cdots+X_n} {\sqrt n} =Y$ for all $n$. so the limiting distribution can be anything. I also believe that there are examples where limit (in distribution) does not exist.