distance from the centre of a $n$-cube as $n \rightarrow \infty$ I've figured out the pattern for calculating the average distance from the centre of an n-cube; but I don't have a formula for the answer. Is there an easy way to figure this out?
Average distance of points from the centre of a unit 0-cube (point)
$$A_0 = 0$$
Average distance of points from the centre of a unit 1-cube (line)
$$A_1 = \int_{x=-\frac{1}{2}}^{x=\frac{1}{2}}{x}\; dx = 0.250000$$
Average distance of points from the centre of a unit 2-cube (square)
$$A_2 = \int_{x=-\frac{1}{2}}^{x=\frac{1}{2}}{\int_{y=-\frac{1}{2}}^{y=\frac{1}{2}}\sqrt{x^2+y^2}}\;dy \; dx \approx 0.382598$$
Average distance of points from the centre of a unit 3-cube (cube)
$$A_3 = \int_{x=-\frac{1}{2}}^{x=\frac{1}{2}}{\int_{y=-\frac{1}{2}}^{y=\frac{1}{2}}\int_{z=-\frac{1}{2}}^{z=\frac{1}{2}}{\sqrt{x^2+y^2+z^2}}}\;dz\;dy \; dx \approx 0.480296$$
Average distance of points from the centre of a unit 4-cube (tesseract)
$$A_4 \approx 0.560950$$
My gut instinct is that $A_n \rightarrow \infty$ as $n \rightarrow \infty$ as in my head higher dimensional cubes become more spiky and I expect the mass to become concentrated in the corners. I feel justified in saying this because the number of "corners" is $2^n$ with a potential distance of $\frac{\sqrt{n}}{2}$ If somehow it were to approach some limit, that would be cool (to me at least)
Thanks in advance for any help, advice or answers
 A: [Note: Added values for five-dimensional hypercube.]
Consider that the space is a hypercube, so each coordinate is independently distributed.  The square of that coordinate's difference from $\frac12$ has the pdf
$$
f(x) = \begin{cases}
\frac{1}{\sqrt{x}} & 0 \leq x \leq \frac14 \\
0 & \text{elsewhere}
\end{cases}
$$
This distribution has a mean of $\frac{1}{12}$ and a variance of $\frac{1}{180}$.  As $n$ increases without bound, the squared distance of the point from the hypercube's center is the sum of $n$ independent and identically distributed (i.i.d.) variables with that same distribution, and is thus asymptotically normally distributed (by the central limit theorem) with mean $\frac{n}{12}$ and variance $\frac{n}{180}$.  For instance, for $n = 180$, we would have a mean squared distance of $15$ and a variance of $1$.  That variance is small enough already that you could just take the square root of the mean squared distance and probably get a very good approximation of the mean distance.
By that logic, the mean distance would be asymptotically $\sqrt{\frac{n}{12}}$, approached from below, since the square root of a nearly normal distribution with a positive mean would be skewed that way.
For $n = 1, 2, 3, 4, 5$, this expression yields approximate mean distances of $0.289, 0.408, 0.500, 0.577, 0.645$, which compares reasonably well with the more accurate values given in the OP ($0.250, 0.383, 0.480, 0.561, 0.631$).  These latter values appear to be approaching the asymptotic expression from below, as expected, but are already not too far off.
ETA ($2019$-$02$-$13$, five-dimensional case added $2020$-$04$-$24$): A second-order analysis yields $\sqrt{\frac{5n-1}{60}}$, for which the values for $n = 1, 2, 3, 4, 5$ are $0.258, 0.387, 0.483, 0.563, 0.632$, respectively, showing even closer agreement.
A: Clearly the limit is infinite as half the volume of the cube lies in the region where the absolute value of the coordinates sum to something greater than n/4.
Ok let's elaborate:
It is enough to consider the positive orthant. Here the map from (x_1,...,x_n) to (1/2-x_1,...,1/2-x_n) shows that exactly half the points by volume have x_1+...+x_n>n/4. The point closest to the origin with this property is (1/4,...,1/4) it is at a distance of sqrt(n)/4 from the origin...
