Let $n\geq 2$ be an integer. We pick $n$ points in $[0,1]$ with uniform distribution. Let $A$ be the minimum distance that two adjacent points have, and let $B$ be the maximum distance that two adjacent point of the points picked have.
Let $X$ be the ratio $B/A$. Does $E(X) \to \infty$ as $n\to \infty$?
$EX=\infty$ 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). 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)) $$f(u,v)=\tfrac12\,(n-1)(n-2)(e^{-u}-e^{-v})^{n-3}e^{-u}e^{-v}\,1_{0<u<v}. $$ So, $$EX=\tfrac12\,(n-1)(n-2)\int_0^\infty dv\int_0^v du \frac vu\,(e^{-u}-e^{-v})^{n-3}e^{-u}e^{-v}=\infty, $$ 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 $v>0$.
