Curse of Dimensionality: hypercube inside a hypersphere Exercise 2.5 of Izenman's Modern Multivariate Statistical Techniques:

Consider a hypercube of dimension $r$ and sides of length $2A$ and
  inscribe in it an $r$-dimensional sphere of radius $A$. Find the
  proportion of the volume of the hypercube that is inside the
  hypersphere, and show that the proportion tends to $0$ as the
  dimensionality $r$ increases. In other words, show that all the
  density sits in the corners of the hypercube.

Let $C$ be the volume of the hypercube, and $S$ be the volume of the hypersphere.
Then
$$\dfrac{C}{S} = \dfrac{(2A)^r}{2\pi^{r/2}A^r/[r\Gamma(r/2)]} = \dfrac{2^{r-1}r\Gamma(r/2)}{\pi^{r/2}}\text{.}$$
Does this really tend to $0$? If so, I don't see it and I don't think this would be true... since (I would think it's obvious that) $r\Gamma(r/2) > \pi^{r/2}$ for large $r$... or am I wrong?

The question is which part of the hypercube volume $C$ is also inside the hypersphere, i.e. in $S\cap C$. Since the hypersphere is fully contained in the hypercube, $S\cap C=S$. So according to Wikipedia you want
$$\frac{V(S)}{V(C)}
=\frac{A^r\frac{\pi^{r/2}}{\Gamma(\frac r2+1)}}{(2A)^r}
=\left(\frac{\sqrt\pi}{2}\right)^r\cdot\frac1{\Gamma(\frac r2+1)}
$$
Now $\frac{\sqrt\pi}2\approx0.886<1$ and $\lim_{x\to+\infty}\Gamma(x)\to+\infty$ so yes, this does tend to zero.
Since $\Gamma\left(\frac r2+1\right)=\frac r2\Gamma\left(\frac r2\right)$ the formula for $V(S)$ agrees with what you used, except that for $r=0$ you'd have $\Gamma(0)$ undefined. But for $r\to\infty$ that's irrelevant.
This whole question reminds me of this post of mine about hypershperes and hypercubes…
