Let $X_j$, $j=1,2,\ldots,X_n$ be independent and take on values $\pm j$ with probability $1/2$ each. Use theorem $2.7.1$ to show that $$\sqrt{\frac3{n}}\bar{X}\xrightarrow{L}N(0,1).$$
Theorem 2.7.1 Liapunov. Let $X_i$ ($i=1,\ldots,n$) be independently distributed with means $E(X_i)=\xi_i$ and variances $\sigma_i^2$, and with finite third moments. If $$Y_n=\frac{\bar{X}-E(\bar{X})}{\sqrt{\operatorname{Var}(\bar{X})}}=\frac{\sqrt{n}\ (\bar{X}-\bar{\xi})}{\sqrt{(\sigma_1^2+\cdots+\sigma_n^2)/n}},$$ then $$Y_n\xrightarrow{L}N(0,1),$$ provided $$\bigg[E\bigg(\sum|X_i-\xi_i|^3\bigg)\bigg]^2=o\bigg[\bigg(\sum\sigma_i^2\bigg)^3\bigg].$$
I am kind of confused as too approach this proof using Theorem 2.7.1. Any help to get me started would be helpful