volume of a cube and a ball in n-dimensions?

Question

Why does the volume of an $n$-dimensional cube approach $\infty$ as $n \to \infty$, but the volume of an $n$-dimensional sphere approach $0$ as $n \to \infty$?

To be more precise,

$$\frac{V_n(\text{Sphere})}{V_n (\text{Cube})} \to 0, \,\text{as}\; n \to \infty$$

I understand how this is evident using the expression for the volume of an $n$-dimensional sphere and cube.

Volume of an $n$-dim sphere (radius = $1$) $$V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}$$

Volume of an $n$-dim cube (side length = $2$) $$V_n = 2^n$$

I am looking for a geometric explanation. In what way is a n-dim sphere different from a n-dim cube? Are there more general families which exhibit these properties?

Edit: For the sake of keeping the problem relatively simple, I am referring to a sphere of unit radius and a cube with side length equal to the diameter of the sphere.

Answer 1

In very high dimensional space, the opposite corners of a cube are very far from one another. The unit cube is a huge object as compared to the unit sphere, where every point is 2 units or less from every other point.

Answer 2

A first remark: For a cube of side $a$ in $n$ dimensions, the volume is $a^n$. Thus the limit depends on whether $a\in(0,1)$, $a=1$ or $a>1$.

You are comparing the volumes $B_n$ of the ball of radius one and $C_n$ of the cube of side two. The reason why $C_n/B_n\to\infty$ as $n\to\infty$ is that, loosely speaking, much of the volume of the cube is near its corners which the ball doesn't reach, and this phenomenon becomes stronger and stronger as dimension grows. In other words, a ball of radius one cannot go very far in every direction at once, but the cube of side two can.