Suppose $X_k$ are independent mean 0 random variables with finite positive variance. Can it be true that $\dfrac{S_n}{\sqrt{\operatorname{Var}(S_n)}}$ does not converge weakly? Here $S_n=\sum_{k=1}^n X_k$.
Almost all the examples I am thinking of, are where the Gaussian limit is not achieved. But is it necessary for the ratio to have a limiting distribution? I guess not, but a hint would be good.
Notice that $\left(\frac{S_{n}}{\sqrt{\operatorname{Var}\left(S_{n}\right)}}\right)_{n\geqslant 1}$ is centered and has variance one, hence this sequence is tight and admits a convergent subsequence. Here the point is that the limit law can be different.
Let $U$ and $V$ be two random variables having variance one and expectation zero; let $\left(Y_i\right)_{i\geqslant 1}$ an independent sequence such that for all $i$, $Y_{2i}$ has the same law as $U$ and $Y_{2i+1}$ the same as $V$. Finally, let $\left(c_i\right)_{i\geqslant 1}$ be a sequence of real numbers such that $$ \frac{\sum_{i=1}^{n-1}c_i^2}{ c_n^2}\to 0 $$ and $X_{i}=c_iY_i$. Then $$ \frac{S_{2n}}{\sqrt{\operatorname{Var}\left(S_{2n}\right)}}\to U\mbox{ and }\frac{S_{2n+1}}{\sqrt{\operatorname{Var}\left(S_{2n+1}\right)}}\to V. $$