Take $X_1 , X_2 , \cdots $ are $iid$ with zero mean. Take $$Z_n = \frac{X_1 + \cdots X_n}{\sqrt{n}} \stackrel{d}{\rightarrow} X$$ and $$Z_{2n} = \frac{X_1 + \cdots X_{2n}}{\sqrt{2n}} \stackrel{d}{\rightarrow} X$$ Name the characteristic function of $X$ to be $f(\xi)$
I managed to show the following: $$f(\xi) = f(\frac{\xi}{\sqrt{2}})^2$$ and that $f(\xi)$ is the characteristic function of a Gaussian when $f \in C^2(\mathbb{R})$
Now I want to show that, if we replace $\frac{1}{\sqrt{n}}$ with $\frac{1}{n}$, then $Z_n$ is distributed Cauchy-Lorentz when $f(\xi) = f(-\xi)$ or $f(\xi) = 1$
I started as so: $$f(\xi) = \mathbb{E}e^{i \xi X} = \mathbb{E}e^{i \xi \frac{X_1 + \cdots X_n}{n}} = \mathbb{E}\prod_{i=1}^{n}e^{i\xi \frac{X_i}{n}}=(\mathbb{E}e^{i\xi \frac{X_1}{n}})^n$$ Then by Taylor expantion $$(\mathbb{E}e^{i\xi \frac{X_1}{n}})^n = \mathbb{E}[1 + \frac{i \xi}{n}X + O(n^{-2})]^n \stackrel{n \rightarrow \infty}{\rightarrow} e^{i\xi (0)} = 1$$
This is however far from a cauchy Lorentz distribution... Where did I go wrong?
Thank you for your insight!
EDIT: I think the reason we cannot use the Strong Law of Large Numbers:
Suppose $\mathbb{E}|X| < \infty $ then $\bar{X_n}$ converges almost surely to $\mathbb{E}X$ In this case, we do not know that $E|X_i| < \infty$ only that $EX_i = 0$
If the iid $X_i$ have zero mean, the sequence $Z_n=(X_1+\dots+X_n)/n$ converges to 0, almost surely. Hence your random variable $X$ is equal to 0 with probability 1, and has the constant function $\phi_X(\xi)=1$ as characteristic function. As is well known, non degenerate Cauchy rvs do not have expected values, so there is something fishy with your problem statement.
Another symptom of hinkiness is your Taylor expansion. If all you know about the rv is that it has a first moment, you are not entitled to either equality in your last displayed equation. You can expand $\exp(i\xi X_1/n)$ in a power series in $\xi$ but you have no warrant for interchanging this summation with $\mathbb E$.