Last time when I derived the formula for the volume of a hypersphere in $n$ dimensions I forgot to point out a curious consequence of the formula, namely that the volume tends to zero as $n$ tends to infinity.

When I was an undergraduate I remember a professor of mine pointing this out and then declaring “That doesn’t make sense!”. At the time it didn’t seem too surprising to me, since I could see that the unit circle in $\newcommand{\R}{\mathbb{R}}\R^2$ took up more of the surrounding square $[-1,1]^2$ than the unit sphere in $\R^3$ took up of $[-1,1]^3$. Consquently, I thought it likely that the ratio of the volume of the unit sphere in $\R^n$ to the volume of $[-1,1]^n$ should go to zero as $n\to\infty$.

However, I misunderstood the claim being made: not only does the above ratio of hypersphere-to-hypercube volume go to zero, the volume of the hypersphere itself goes to zero. This was something I hadn’t even considered: since as $n\to\infty$ the hypersphere is “growing”, I presumably took for granted that its volume should go to infinity, not zero!

Of course, one can consider the unit sphere in $\R^{n-1}$ as a subset of the unit sphere in $\R^n$, since for example the unit sphere in $\R^3$ contains the unit circle as a “slice”. In this way as $n\to\infty$ the hypersphere is growing. However, though the “slice” has volume in $\R^{n-1}$, it has no volume in $\R^n$; as the dimension increases it becomes “harder” to make volume in a sense. This allows the hypersphere to “grow” as $n\to\infty$ while still shrink in volume.

Algebraically, as we’ve seen, the volume of the unit sphere in $\R^n$ is given by

$$V_n = \frac{\pi^{n/2}}{(n/2)!} .$$

If one knows Stirling’s approximation

$$n! \sim \sqrt{2\pi n}\Bigl(\frac{n}{e}\Bigr)^n$$

then it isn’t too hard to see that the denominator of $V_n$ grows asymptotically faster than the numerator, and therefore $V_n$ tends to $0$. Explicitly, we have

$$\lim_{n\to\infty} V_n = \lim_{n\to\infty}\frac{\pi^{n/2}}{\sqrt{\pi n}(\frac{n}{2e})^{n/2}} = \lim_{n\to\infty}\frac{1}{\sqrt{\pi n}}\cdot\lim_{n\to\infty}\Bigl(\frac{2\pi e}{n}\Bigr)^{n/2} = 0$$

since $\lim_{n\to\infty}1/\sqrt{\pi n}=0$ and

$$\lim_{n\to\infty}\Bigl(\frac{2\pi e}{n}\Bigr)^{n/2} = \lim_{n\to\infty}\exp\Bigl(\frac{n}{2}\ln\Bigl(\frac{2\pi e}{n}\Bigr)\Bigr) = \lim_{m\to-\infty}\exp(m) = 0 .$$