Given that $X_i$ are symmetric about 0 and iid with $E[|X_i|]=\infty$. Show that $\frac{S_n}{n}=\frac{X_1+\dots+X_n}{n}$ does not converge to 0. I've been trying something like..
Let us assume that $\frac{S_n}{n}$ converges to 0. That means that,
$$\frac{S_{n+1}}{n+1}-\frac{S_n}{n}$$ converges to 0 too. But we can rewrite this as 
$$\frac{nX_{n+1}}{n(n+1)}-\frac{S_n}{n(n+1)}=\frac{X_{n+1}}{n+1}-\frac{1}{n+1}\cdot\frac{S_n}{n}$$
But from our assumption, the negative part of the above converges to 0, so we are left with 
$$\frac{X_{n+1}}{n+1}$$
and one would think that this should also converge to 0, but because $X_i$ does not have finite first moments (i.e. because $E[|X_i|]=\infty$), we cannot make such a claim so we have reached a contradiction, and thus our initial assumption is false, and $\frac{S_n}{n}$ does not converge to 0.
 A: You are precisely right if by $convergence$ you mean almost sure convergence. If $n^{-1}S_n\to 0$ almost surely, then $\operatorname E|X|<\infty$, because $n^{-1}X_n\to0$ almost surely if and only if $\operatorname E|X|<\infty$ (see this question).
However, the answer to your question depends on the type of convergence. If we investigate convergence in probability, then $n^{-1}S_n$ converges to $0$ if and only if $x\Pr\{|X|>x\}\to0$ as $x\to\infty$ and this is a weaker condition than $\operatorname E|X|<\infty$. If we investigate almost sure convergence, then $n^{-1}S_n$ converges to $0$ if and only if $\operatorname E|X|<\infty$.
A: Please disregard this response. I had gotten confused while discussing this outside of M.SE with the asker of the question. My interpretation was that it should be disproved that $S_n/n$ converges to 0, which is not the same as what is written. 
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We provide a counterexample. Let $X_1, X_2, \ldots \sim F = \text{Cauchy}(0,1)$. Then, $$f_{X_i}(x) = \frac{1}{\pi (1+x^2)},$$ from which it follows that $f_{X_i}(x) = f_{X_i}(-x) \forall x \in \mathbb{R}$ and  $$\mathbb{E}|X_i| = 2\int_0^\infty \! \frac{x}{\pi(1+x^2)} \, \mathrm{d}x = + \infty.$$
Cauchy(0,1) random variables $X,Y$ have the properties $X+Y \sim \text{Cauchy}(0,2)$ and $kX \sim \text{Cauchy}(0,|k|)$ for $k \in \mathbb{R}$. Hence, $n^{-1}\sum_{i=1}^n X_i \sim \text{Cauchy}(0,n/n) = \text{Cauchy}(0,1), \forall n \in \mathbb{N}$. It's obvious then that $$P\left(\left|\frac{S_n}{n}\right| > \epsilon\right) = c > 0, \forall n \in \mathbb{N}.$$ Therefore, $$\lim_{n \to \infty} P\left(\left|\frac{S_n}{n}\right| > \epsilon\right) = c > 0.$$
