Question-
$X_n$ can take only two values $n^a$ and $-n^a$ with equal probabilities. Show that we can apply weak law of large numbers to the sequence of independent random vatiables ${X_n}$ if $a<\frac{1}{2}$.
We have to show that$Var(\overline{X_n})$ $\to 0$ as $n\to\infty$. I can show that if $a>1/2$ then $Var(\overline{X_n})$ does not tend to $0$ but i can not prove that wlln can be applied if $a<1/2$. Any help would be appreciated!
Since you don't provide us with any details on your calculations, it is hard to say where you went wrong.
Since the random variables $X_n$, $n \geq 1$, are independent and have mean $0$, it holds for $\bar{X}_n := n^{-1} \sum_{i=1}^n X_i$ that
$$\text{var}(\bar{X}_n) = \frac{1}{n^2} \sum_{i=1}^n \text{var}(X_i) = \frac{1}{n^2} \sum_{i=1}^n \mathbb{E}(X_i^2).$$
By assumption,
$$\mathbb{E}(X_i^2) = \frac{1}{2} (i^a)^2 +\frac{1}{2} (-i^a)^2 = i^{2a},$$
and so
$$\text{var}(\bar{X}_n)= \frac{1}{n^2} \sum_{i=1}^n i^{2a}.$$
Since $i^{2a} \leq n^{2a}$ for any $i \in \{1,\ldots,n\}$ this implies
$$\text{var}(\bar{X}_n)\leq \frac{n^{2a}}{n} \xrightarrow[2a<1]{n \to \infty} 0.$$
Applying Markov's inequality (/Tschebysheff inequality) it follows that the weak law of large numbers holds.