The \(n\)-dimensional unit sphere (in the indexing in which the Earth is a 3-dimensional sphere) has an inner volume which is its surface area divided by \(n\) (by considering each tiny patch of its surface as the base of a figure tapering down to its center), and at the same time, by the argument that projecting two dimensions outwards to the enveloping “cylinder” is area-preserving (stretching latitudinally and longitudinally in precise cancellation, as in the Lambert map projection), we find that the unit \(n\)-sphere’s surface area is \(\tau\) times the unit \((n - 2)\)-sphere’s volume (\(\tau\) here being the circumference of a unit circle).1

Putting these last two facts together, the unit \(n\)-sphere’s volume is \(\frac{\tau}{n}\) times the unit \((n - 2)\)-sphere’s volume.

Letting \(V(n)\) be the unit \(n\)-sphere’s volume, and letting \(D(n) = \frac{V(n + 1)}{V(n)}\), we get that \(\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \ldots \times \frac{n}{n + 1}\) comes out to \(\frac{V(0)/V(n)}{V(1) / V(n + 1)} = \frac{V(0)}{V(1)} D(n) = \frac{1}{2} D(n)\). And of course \(\frac{2}{1} \times \frac{4}{3} \times \frac{6}{5} \times \ldots \frac{n}{n - 1}\), with as many factors, is just this same thing times \(n + 1\). Putting these together, the Wallis product up through \(\frac{n}{n + 1} \times \frac{n}{n - 1}\) is \(\frac{n + 1}{4} D(n)^2\).

But what are the asymptotics of that \(D(n)^2\) factor? Well, our recurrence \(V(n) = \frac{\tau}{n} \times V(n - 2)\) is equivalently the recurrence \(D(n - 2) \times D(n - 1) = \frac{\tau}{n}\). So both \(D(n - 1) \times D(n)\) and \(D(n) \times D(n + 1)\) are \(\sim \frac{\tau}{n}\). If we just knew that \(D(n)\) was monotonic in \(n\), then \(D(n)^2\) would be between those two and thus \(\sim \frac{\tau}{n}\) itself. Given our previous paragraph’s conclusion, the Wallis product would then come out to \(\frac{\tau}{4}\) (i.e., \(\frac{\pi}{2}\)). Ta-da!

…Well, one lemma to establish, and then we can ta-da. Let’s see why \(D(n)\) is monotonic in \(n\):

Considering the \((n + 1)\)-sphere’s volume in terms of its cross-sectional \(n\)-spheres, we have that \(D(n)\) is twice the average value of \(f(x)^n\) as \(x\) runs from \(0\) to \(1\), where \(f(x) = \sqrt{1 - x^2}\) is the cross-sectional radius at height \(x\) above the center of a unit sphere. And since \(f(x)^n\) is a decreasing function of \(n\) (since \(f(x) \in [0, 1]\)), thus so is \(D(n)\).

Now we are done. Ta-da!

This is actually related to Wallis’s original proof (which was done not in terms of spheres but in terms of some messy integrals; even messier perhaps by virtue of being worked out before the general integral calculus had been developed!), but that is as this seen through a glass very darkly. I would be very curious to see if anyone has written on seeing the Wallis product this way in terms of sphere volumes before.

Footnotes:

  1. Here, whenever I say “(inner) volume”, I mean in the sense appropriate to the number of dimensions, and similarly for “surface area”; thus, for example the “surface area” of a circle is its perimeter, and the “volume” of a circle is its area. ↩