On the largest and smallest spacings for the uniform distribution Let $Z_1,\dots,Z_n$ be iid random variables (r.v.'s) each uniformly distributed on $[0,1]$. Let $Z_{n:1}\le\cdots\le Z_{n:n}$ be the corresponding order statistics. For $i=1,\dots,n-1$, let $G_i:=Z_{n:i+1}-Z_{n:i}$, the $i$th spacing/gap. Let 
\begin{equation}
 A_n:=G_{n-1:1}=\min_{i\le n-1}G_i,\quad B_n:=G_{n-1:n-1}=\max_{i\le n-1}G_i. 
\end{equation}
In comments at Expected value ..., Brendan McKay asked if $EB_n/EA_n\to\infty$ (as $n\to\infty$), and Anthony Quas asked if $med(B_n/A_n)\to\infty$, where $med$ denotes the median. 
The purpose here is to answer these questions (affirmatively). 
 A: $\newcommand{\eD}{\overset{\text{D}}\to} 
\newcommand{\D}{\overset{\text{D}}=}$ 
As was noted on the linked MO page Expected value ..., 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 (r.v.'s); see e.g. Theorem 6.6(c). 
So,
\begin{equation*}
B_n\D M_n:=\frac{H_{n-1:n-1}}{S_{n+1}}=\frac1{S_{n+1}}\,\max_{i\le n-1}H_i, 
\end{equation*}
where $\D$ means the equality  in distribution, and 
$S_{n+1}:=H_1+\dots+H_{n+1}$. 
Next, for any real $x$ and all large enough natural $n$,
\begin{multline*}
 P(H_{n-1:n-1}-\ln n<x)=P(\max_{i\le n-1}H_i<x+\ln n)  
 =P(H_1<x+\ln n)^{n-1} \\ 
 =(1-e^{-x-\ln n})^{n-1}
 \to e^{-e^{-x}}=P(Y<x)
\end{multline*}
for some r.v. $Y$, so that 
\begin{equation*}
 Y_n:=H_{n-1:n-1}-\ln n\eD Y, 
\end{equation*}
where $\eD$ means the convergence in distribution. 
Also, by the strong law of large numbers (SLLN) $\frac n{S_{n+1}}\to1$ almost surely and hence in distribution. 
So, 
\begin{equation*}
 \frac n{\ln n}\,B_n\D\frac n{\ln n}\,M_n=\frac n{\ln n}\,\frac{H_{n-1:n-1}}{S_{n+1}}
 =\frac{Y_n+\ln n}{\ln n}\,\frac n{S_{n+1}}\eD1. \tag{1}
\end{equation*}
So, by the Fatou lemma, 
\begin{equation*}
 \liminf_n\frac n{\ln n}\,EB_n\ge1. 
\end{equation*}
On the other hand,
\begin{equation*}
 A_n\le G_1,
\end{equation*}
and $G_1$ has the beta distribution with parameters $1,n$. So,
\begin{equation*}
 EA_n\le EG_1=\frac1{n+1}. 
\end{equation*}
So, 
\begin{equation*}
 \liminf_n\frac{EB_n}{EA_n}\ge\lim_n\frac{\ln n}n\,(n+1)=\infty. 
\end{equation*}
Thus, it is confirmed that $EB_n/EA_n\to\infty$. 
Also, 
\begin{equation}
 \frac{B_n}{A_n}\ge \frac{B_n}{G_1}
 \D\frac{nB_n}{H_1}\,\frac{S_{n+1}}n\eD\infty, 
\end{equation}
because, by (1), $nB_n\eD\infty$ and, by the SLLN, $\frac{S_{n+1}}n\eD1$. 
Thus, $\frac{B_n}{A_n}\eD\infty$ and hence indeed $med(B_n/A_n)\to\infty$. 
