# Tag Info

If one uses the sphere volume as $\pi^{n/2}/\frac n2 !$ and then apply stirling's approximation to $n! = (n/e)^n$, then one gets $(2.\pi.e/n)^{n/2}$ as the approximation for the volume of an n-sphere. This then equates to a cube of edge $\sqrt{2\pi e/n}$. Of course, one can use by way of limits, that even a cube of side $1/2$ must 'go to zero', even ...