So, I was watching this YouTube video about pi and the Sierpinski Carpet/Menger Sponge, which talks about this interesting fact:

Say you build a Sierpinski Carpet, starting with a 2x2 square, but instead of always cutting the square into 3x3 and taking out the centre square, you cut it into successive odd numbers and take out the middle square. That is, for the first level you cut each square into 3x3 and take out the centre 1, then for the next level you cut each of the 8 squares into 5x5 and take out the centre 1, and so on. Then the area of the final carpet no longer converges to zero... you get the product of (8/9) * (24/25) * (48/49) * ..., which you can massage into the Wallis product, and you end up with an area of pi. In particular, you end up that this carpet has the same area as a unit circle inscribed in the original square. Apparently this is also called the Wallis sieve.

Go up a level. Say you build a Menger Sponge in the same way. You start with a 2x2x2 cube, and cut it into 3x3x3, remove the middle axes, then cut each of the remaining 20 cubes into 5x5x5 and remove the middle axes, and so on. Now, I don't know enough about solving infinite products to really understand why, but the video references an article called "Squeezing Pi from a Menger Sponge" that says this converges to a volume of 4/3pi... the same volume as a unit sphere inscribed in the original cube.

Now, after watching this video, I was curious if it continued to higher dimensions. Now, I don't have the knowhow to actually solve any of these infinite products, but I do have Wolfram|Alpha and a list of n-ball volume formulae from Wikipedia...

If you take a 4D hypercube, and divide it into 3x3x3x3, punch out the middles (specifically, remove any subhypercube that is in the middle third on at least two axes), you're left with 48/81. Then divide it into 5x5x5x5, punch out the middles, and you're left with 512/625. Then continue.

For step i, you're dividing it into (2i+1)

^{4}cubes, and keeping (16i

^{4}+ 32i

^{3}) of them. I have no idea how you derive it, but if you take that infinite product, multiply it by 16, and plug it into Wolfram|Alpha, you get pi

^{2}/2. Which is exactly the volume of the unit 4-ball.

To continue even higher:

For a n-dimensional Menger sponge, start with a 2

^{n}hypercube. Then, for the i-th term, you're dividing each cube into (2i+1)

^{n}subcubes, and then you're keeping the ones which are either (a) not in the centre on any axis, or (b) in the centre on exactly one axis. In a sense, these are the corners and the edges of the sponge.

For part (a), this is simple enough - on each axis, there are 2i coordinates that aren't in the centre, so this covers (2i)

^{n}cubes. For part (b), there are n dimensions that can be in the centre, and (2i)

^{n-1}values for the other coordinates, for a total of n(2i)

^{n-1}cubes.

So, putting it all together, the question is: is

2

^{n}* product i=1 to infinity of ((2i)

^{n}+ n(2i)

^{n-1}) / (2i+1)

^{n}

equal to the area of the unit n-sphere?

Wolfram|Alpha claims it's correct at least as far as n=8... beyond that, the calculation times out. But does it hold for all n? And can you prove it? For that matter, can you prove it for these specific cases, in a way that doesn't involve "and then I plugged it into W|A and then magic happened"?