I am working with the Cauchy random variable. I have a sequence of iid Cauchy r.v. $X_i$ with parameters $0,\lambda>0$, i.e. $\phi_{X_1}(t)=e^{-\lambda|t|}$.
I think I have shown that $\frac{S_n}{n} \to X_1$ in distribution. Here's the proof, using the charateristic functions: $\phi_{\frac{S_n}{n}}(t)=\mathbb{E}(e^{it\frac{S_n}{n}}) = \mathbb{E}(e^{\frac{it}{n}(X_1+\ldots+X_n)}) = \mathbb{E}(\prod_{j=1}^n e^{\frac{it}{n}X_j}) = \prod_{j=1}^n(\mathbb{E}(e^{\frac{it}{n}X_j})) = \mathbb{E}(e^{\frac{it}{n}X_1})^n = \exp(-\lambda|\frac{t}{n}|)^n= \exp(-\lambda|t|)=\phi_{X_1}(t)$
I am trying to prove now that $S_n/n$ does not converge to $X_1$ in probability. I am trying to do so by contradiction, supposing it does and then proving $S_n/n - S_{2n}/2n $ does not converge in distribution to $0$ (thus contradicting the first statement). But I did not manage to do so because every time I extended the computation it did go to the characteristic function of $0$ (if you have any suggestion in this direction they will be very well accepted). Since my problem was about the factorization of the $2$ and $n$ I started trying with $S_{2n}/n - 2S_n/n$ which should go to $0$ in probability and I seemed to reach a contradiction, but I am not sure it works formally.
I think you are expected to show that $S_n/n$ does not converge to any random variable in probability. Here is a proof: if $S_n/n$ converges in probability then $S_n/n-S_{2n}/\{2n\} \to 0$ in probability. Since convergence in probability implies convergence in distribution we must have $Ee^{it (S_n/n-S_{2n}/\{2n\})} \to 1$ for each $t$. The characteristic function is $e^{-\lambda |t|}$ for each $n$, so we have a contradiction.