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 $n\to \infty$?



Expected value of "longest bit / shortest bit" in $n$ uniformly distributed points on $[0,1]$
$EX=\infty$ 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)](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). 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)](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))
$$f(u,v)=\tfrac12\,(n-1)(n-2)(e^{-u}-e^{-v})^{n-3}e^{-u}e^{-v}\,1\_{0<u<v}. 
$$
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 $v>0$. 


