$EX=\infty$ for all $n\ge3$. Indeed, the gaps $G_1,\dots,G_{n-1}$ between the adjacent points are jointly distributed as $\frac{H_1}{H_1+\dots+H_{n+1}},\dots,\frac{H_{n-1}}{H_1+\dots+H_{n+1}}$, where the $H_i$'s are iid standard exponential random variables; see e.g. [Theorem 6.6(c)][1]. So, $X=B/A$ equals $V/U$ in distribution, where $V:=\max_{i\le n-1}H_i$ and $U:=\min_{i\le n-1}H_i$. The joint pdf of $(U,V)$ is (see e.g. [Theorem 6.2(e)][1])
$$f(u,v)=\tfrac12\,(n-1)(n-2)(e^{-u}-e^{-v})^{n-3}e^{-u}e^{-v}\,1_{0<u<v}. 
$$
So, 
$$EX=(n-1)(n-2)\int_0^\infty dv\int_0^v du \frac vu\,(e^{-u}-e^{-v})^{n-3}e^{-u}e^{-v}=\infty, 
$$
because 
$$\int_0^v du \frac1u\,(e^{-u}-e^{-v})^{n-3}e^{-u}
\ge\Big(\int_0^{v/2} \frac{du}u\Big)\,(e^{-v/2}-e^{-v})^{n-3}e^{-v/2}=\infty
$$
for all real $v>0$. 


  [1]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=2ahUKEwjmvI-6vp7kAhUDiqwKHQnXBN4QFjABegQICRAE&url=http%3A%2F%2Fwww.stat.purdue.edu%2F~dasgupta%2Forderstats.pdf&usg=AOvVaw27BT2T6w-qpihgz3e69jaW