Does the proportion of the volume of the hypercube to the volume of the containing hypersphere tend to 1 as dimensions grow?

Question

I read on this website that an "infinite dimensional" (limiting case of dimensionality) hypercube containing a hypersphere has a ratio of the hypersphere:hypercube volumes that tends to 0. Is it also true that a hypersphere containing a hypercube will have a ratio of volumes such that they tend to 1? Containing is meant to imply that the hypercube is inscribed inside of the hypersphere. Note that all vertices of the hypercube lie on the hypersphere.

Answer 1

The volume of a hyperball of radius $R$ is $$V_{\rm ball} = \frac{\pi^{n / 2}}{\Gamma\left(\frac{n}{2} + 1\right)} R^n,$$ and the volume of a hypercube of side length $s$ is $s^n$. A straightforward geometric argument shows that the side length of a hypercube inscribed in a hyperball of radius $R$ in dimension $n$ is $s = \frac{2 R}{\sqrt{n}}$, so the inscribed hypercube has volume $$V_{\rm cube} = \frac{2^n R^n}{\sqrt{n}} .$$ The ratio of volumes is thus $$\frac{V_{\rm cube}}{V_{\rm ball}} = \left(\frac{4}{\pi n}\right)^{n / 2} \Gamma\left(\frac{n}{2} + 1\right) ,$$ which tends to zero as $n \to \infty$.

We can be more specific about the rate of decay of the ratio: Stirling's Approximation gives that $$\Gamma(x + 1) \approx \sqrt{2 \pi x} \left(\frac{x}{e}\right)^x ,$$ and substituting yields $$\frac{V_{\rm cube}}{V_{\rm ball}} \approx \frac{1}{\sqrt{\pi n}} \left(\frac{2}{\pi e}\right)^{n / 2} .$$