Suppose $X_1,X_2,...$ are i.i.d with $\mathcal{N}(0,1)$. Denote $S_n=X_1+...+X_n$. I am trying to determine if the sequence of random variables defined as: $$\frac{S_n}{\sqrt{n}} -\frac{S_{2n}}{\sqrt{2n}}$$ converges in distribution and if so, what is the limiting distribution. Here is my thinking so far of which I tend to be doubtful. Rewriting the above I get: $$\frac{X_1+...+X_n}{\sqrt{n}}-\frac{X_1+...+X_n}{\sqrt{2n}}-\frac{X_{n+1}+...+X_{2n}}{\sqrt{2n}}=$$ $$\frac{(1-\frac{1}{\sqrt{2}})(X_1+...+X_n)}{\sqrt{n}} -\frac{(\frac{1}{\sqrt{2}})(X_{n+1}+...+X_{2n})}{\sqrt{n}}$$
If I start taking large $n$, then by the central limit theorem the distributions of both rv's $\frac{X_1+...+X_n}{\sqrt{n}}$ and $\frac{X_{n+1}+...+X_{2n}}{\sqrt{n}}$ will be closer and closer to $\mathcal{N}(0,1)$. Furthermore, for any fixed $n$ they are independent rv's. So for large $n$, I am effectively dealing with a linear combination of two independent $\mathcal{N}(0,1)$ rv's. Hence, I get $$\frac{(1-\frac{1}{\sqrt{2}})(X_1+...+X_n)}{\sqrt{n}} -\frac{(\frac{1}{\sqrt{2}})(X_{n+1}+...+X_{2n})}{\sqrt{n}}\sim\mathcal{N}(0,(1-\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2)$$ and consequently: $$\frac{S_n}{\sqrt{n}} -\frac{S_{2n}}{\sqrt{2n}}\xrightarrow{d} \mathcal{N}(0,(1-\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2) $$
I am aware that this is not a rigorous proof, but is my intuition correct or is it misleading me completely? Will highly appreciate any help with this problem.