Background Knowledge:
Central Limit Theorem: Assume that $X_1,\dots,X_n$ are i.i.d random variables with mean $\mu$ and variance $\sigma^2$. >Then, $\lim_{n\to\infty} \sqrt{n}(\frac{\bar{X}_n-\mu}{\sigma})$ is the standard normal distribution (where $\bar{X}_n$ is the sample mean). (taken from Wikipedia)
Question:
Assume that $X_1,X_2,\dots,X_n$ are i.i.d random variables having mean $0$ and variance $\sigma^2$. Define:
$$S_n:=X_1+X_2+\dots+X_n$$ $$Y_n:=\frac{S_n}{\sigma\sqrt{n}}-\frac{S_{2n}}{\sigma\sqrt{2n}}$$
Using the central limit theorem, find the value of $\lim_{n\to\infty}Y_n$.
My try:
Applying the central limit theorem, we have: $\lim_{n\to\infty} \sqrt{n}(\frac{\bar{X}_n}{\sigma})$ is the normal distribution. I do not get it. How should I reach for calculating the actual limit in the question?!
Finding the limit of $\frac{S_n}{\sigma\sqrt{n}}-\frac{S_{2n}}{\sigma\sqrt{2n}}$ using the central limit theorem
We can write $$ \frac{S_n}{\sigma\sqrt{n}}- \frac{S_{2n}}{\sigma\sqrt{2n}}=-\frac{1}{\sqrt{2}}\frac{1}{\sigma\sqrt{n}}\sum_{i=n+1}^{2n} X_i + \left(1-\frac{1}{\sqrt{2}}\right)\frac{1}{\sigma\sqrt{n}}\sum_{i=1}^n X_i.$$ Let $$Y_n:=\frac{1}{\sigma\sqrt{n}}\sum_{i=n+1}^{2n}X_i$$ and $$Z_n:=\frac{1}{\sigma\sqrt{n}}\sum_{i=1}^n X_i.$$ These are independent and converge in distribution to standard normal by the CLT. So what does $$ -\frac{1}{\sqrt{2}} Y_n + \left(1-\frac{1}{\sqrt 2}\right) Z_n$$ converge in distribution to?