Assuming that $\{X_n\}$ is a sequence of independent random variables, and $P\{X_n=\pm n^\alpha\}=\frac{1}{2}$, $n=1,2,\cdots$.
It can be easily proved that if $0<\alpha<\frac{1}{2}$, $\{X_n\}$ satisfies the weak law of large numbers, i.e. $\frac{1}{n}\sum_{k=1}^{n}X_k\xrightarrow{P}0$, as $n\rightarrow\infty$, with the Chebyshev's inequality.
But I have no idea how to prove that if $\alpha\geq\frac{1}{2}$, the weak law of large numbers is not satisfied. Anyone can help?
$S_n/n=\frac{1}{n}\sum_{k=1}^{n}X_k\xrightarrow{P}0$ iff characteristic function of this average converges pointwise to $1$: $$ \varphi_{S_n/n}(t)\to 1 \text{ as } n\to\infty. $$ Find it. $$ \varphi_{S_n/n}(t) = \varphi_{X_1}(t/n)\cdot\ldots\cdot \varphi_{X_n}(t/n) = \cos\left(\frac{t}{n}\right)\cos\left(\frac{2^\alpha t}{n}\right)\cdot\ldots\cdot \cos\left(\frac{n^\alpha t}{n}\right). $$ Let $\alpha\geq 1$. Then the product does not converges to $1$ since absolute value of all the factors do not exceed $1$, and last factor $\cos(n^{\alpha-1}t)$ does not converges to $1$.
Let $\frac12\leq \alpha <1$. Then all the factors are positive for $0\leq t<\pi/2$, and $\cos(t)$ is decreasing function inside this interval. We can take $n$ even and bound the product from above as: $$ \varphi_{S_n/n}(t) =\cos\left(\frac{t}{n}\right)\cos\left(\frac{2^\alpha t}{n}\right)\cdot\ldots\cdot \cos\left(\frac{n^\alpha t}{n}\right)\leq \left[\cos\left(\frac{\left(\frac{n}2\right)^\alpha t}{n}\right) \right]^{n/2}=\left[\cos\left(\frac{n^{\alpha-1} t}{2^\alpha}\right) \right]^{n/2}. $$ Indeed, we bound the first $n/2$ of factors by $1$ from above, and the other $n/2$ factors $\cos\left(\frac{k^\alpha t}{n}\right)$, $\frac{n}2<k\leq n$ - by cosine of smaller value $\frac{\left(\frac{n}2\right)^\alpha t}{n}$. We can also bound the characteristic function from below by the $n$th power of the last smallest factor: $$\varphi_{S_n/n}(t) =\cos\left(\frac{t}{n}\right)\cos\left(\frac{2^\alpha t}{n}\right)\cdot\ldots\cdot \cos\left(\frac{n^\alpha t}{n}\right)\geq \left[\cos\left(\frac{n^\alpha t}{n}\right) \right]^{n}=\left[\cos\left(n^{\alpha-1} t\right) \right]^{n}. $$ Note that $\cos\left(n^{\alpha-1} t\right)=1- \frac{n^{2\alpha-2}t^2}{2}+o(n^{2\alpha-2}t^2)$ as $n^{2\alpha-2}t^2\to 0$ since $2\alpha-2<0$. So the bounds from below and from above are asymptotically equivalent to $$ \left(1-\frac{n^{2\alpha-2}t^2}{2}\right)^n \text{ and } \left(1-\frac{n^{2\alpha-2}t^2}{2^{2\alpha+1}}\right)^{n/2} $$ Both this values converge to $1$ only if $n\cdot n^{2\alpha-2}\to 0$ as $n\to \infty$, but this is not the case. For $\alpha=1/2$ the above bound converges to $e^{-t^2/8}$, and for $1/2<\alpha<1$ it converges to $0$.