What is the limit, as $n \to \infty$, of the expected distance between two points chosen uniformly at random within a unit edge-length hypercube in $\mathbb{R}^n$?
For $n=1$, the average distance is $\frac{1}{3}$. For $n=2$, it is approximately $0.52$. For $n=3$, approximately $0.66$ (Robbins' constant).
But I have not found an expression for arbitrary $n$...
The limit is $\infty$. See equation (2) of http://mathworld.wolfram.com/HypercubeLinePicking.html
I think this easy probability argument, using the natural coupling, gives us the limit and the growth rate.
Let $\{U_i, V_i, i=1,2,\dots\}$ be iid uniform on $[0,1]$. Let $$D_n := \sqrt{(U_1 - V_1)^2 + \dots + (U_n - V_n)^2}$$ so that $E D_n$ is the expected distance between two independently and uniformly chosen points in $[0,1]^n$.
By the strong law of large numbers, $\frac{D_n^2}{n} \to E D_1^2 = \frac{1}{6}$ almost surely. Hence $\frac{D_n}{\sqrt{n}} \to \frac{1}{\sqrt{6}}$ almost surely. Moreover, since $E\frac{D_n^2}{n} = \frac{1}{6}$ for all $n$, $\frac{D_n}{\sqrt{n}}$ is bounded in $L^2$ and hence uniformly integrable. So we therefore have $$\frac{1}{\sqrt{n}} E D_n \to \frac{1}{\sqrt{6}}.$$ That is, $E D_n$ goes to infinity like $\sqrt{\frac{n}{6}}$.
