$X_1, X_2,...$ be i.i.d. with characteristic funciton $\phi(t) = e^{-|t|^\alpha}$, $\alpha < 1$. Show that the weak law of large numbers does not hold Let $X_1, X_2,...$ be i.i.d. with characteristic funciton $\phi(t) = e^{-|t|^\alpha}$, where $0 < \alpha < 2$. Suppose $\alpha < 1$. Show that the weak law of large numbers does not hold, that is, there is no constant $\mu$ such that $\frac{X_1+\cdots + X_n}{n} \to \mu$ in probability.
If it is of any use, I have already showed that $\frac{X_1+\cdots + X_n}{n^\frac{1}{\alpha}}$ has the same distribution as $X_1$, and $var(X_1) = \infty$. But I'm still unsure how to approach this problem.
 A: @nejimban already provided a solution based on the actual hypothesis of the exercise. Here is another one in which the infinitely divisible assumption is dropped.
Theorem (Theorem 4 of Feller's An Introduction to probability Theory and Its Applications, vol 2, sect. VII.9, p. 241). Let $(X_n)_n$ be a sequence if independent variates with a common law satisfying $\mathbf{E}(|X_1|) = \infty.$ For whatever the constant $c_n$ may be,
$$
\limsup |n^{-1} S_n - c_n| = \infty,
$$
with probability one, where $S_n = X_1 + \ldots + X_n.$
Sketch of proof. Let $A_k(m) = \{|X_k| > mk\}.$ Then, the events $A_k(m)$ are independent and $\sum\limits_{k = 1}^\infty \mathbf{P}(A_k(m)) = \infty,$ the Borel-Cantelli lemma implies that infinitely many of them occur almost surely. Therefore, the event $L(m) = \left\{\limsup\limits_{k \to \infty} \dfrac{|X_k|}{k} \geq m \right\}$ has probability one, and so $\bigcap\limits_{m = 1}^\infty L(m)$ also has probability one. As a consequence, $\dfrac{|X_k|}{k}$ is an unbounded sequence a.s. Since $X_k = S_k - S_{k-1},$ the boundedness of $\dfrac{|S_n|}{n}$ would entail that of $\dfrac{|X_k|}{k}$ and so $\dfrac{|S_n|}{n}$ is unbounded, proving the claim for $c_n = 0.$ The proof of the general case follows by a "symmetrisation" argument, meaning we consider $Y_k = X_k - X_k',$ where $(X_n')_n$ is an independent copy of the whole sequence $(X_n).$ It can be shown that the $Y_k$ are independent, identically distributed with $\mathbf{E}(|Y_1|) = \infty.$ A fortiori, $\dfrac{|S_n^Y|}{n}$ is unbounded (with obvious choice of notation), and $S_n^Y$ is a symmetrisation of $S_n - n c_n,$ and the boundedness of $\dfrac{|S_n - n c_n|}{n}$ would entail that of $\dfrac{|S_n^Y|}{n},$ which then implies that $\dfrac{|S_n-nc_n|}{n} = |n^{-1}S_n - c_n|$ is unbounded, as desired. CQFD
