# An infinite sum formula for pi

/My fabulous podcast cohost, Evelyn Lamb, is competing in something called the Big Internet Mathoff. Her first entry is about the Wallis sieve and you can read it here. The summary is the following. Take a square of side length 1. Divide each side in thirds and remove the middle square. If you wanted to construct the Sierpinski carpet, you would iterate this procedure, dividing each of the remaining 8 squares into ninths and removing the middle square of each. The Wallis sieve does something different. After removing the middle third square, divide each of the remaining 8 squares into 25 pieces (a \(5\times 5\) grid) and remove the middle square of each of those. Then divide the squares remaining into 49 pieces and remove the middle squares. And so on. What remains is the Wallis sieve.

What is the area of this object? When you build the Sierpinski carpet, it turns out that you remove everything (well not really, but the area is 0). That's not true with the Wallis sieve. Thanks to the Wallis product formula we can see that the area of the sieve is \(\pi/4\). Cool.

But I want to look at this from a different point of view: let's add up the areas of the pieces we remove. Since the area of the Wallis sieve is \(\pi/4\) and we started with a square of area 1, the areas of the pieces we removed must add up to \(1-\pi/4 \approx 0.21460183\dots\) What does this infinite series look like? The first term is \(1/9\), the area of the small square in the center. Then for each of the 8 remaining squares we remove a square of area \(1/(3\cdot 5)^2 = 1/225\). Then in each of those 8 squares we divide the 24 remaining squares into a \(7\times 7\) grid and remove the center square; the next term in the series is then \(8\cdot 24/(3\cdot 5\cdot 7)^2\). Do you see the pattern? The denominators are the squares of the double factorials \((2n+1)!! = 3\cdot 5\cdot 7\cdot (2n+1)\). The numerators are \(1\cdot 8\cdot 24\cdot 48 \cdots (4n)(n-1)\) (you might need to work that out on a piece of paper). There are lots of factors that can be rearranged in that numerator. I'll leave it to you to work out the details, but the series we end up with is \[\sum_{n=1}^\infty \frac{4^{n-1}n!(n-1)!}{[(2n+1)!!]^2} = 1-\frac{\pi}{4}.\]

Cool! This is the sort of series I might put on a quiz in Calculus II as a question about the ratio test. All those factorials just scream for using this test to determine convergence. For this series, though, the test is inconclusive, yielding a ratio of 1. None of the other standard tests we teach our students will show that this series converges, but thanks to Evelyn's post, we know that in fact it *does* converge and we even know its sum.

Given that this series hangs on the edge of the ratio test, one might wonder how quickly the series converges. Well, I started adding up terms. The individual terms do go to 0 in the end, but they're not in much of a hurry to do so. After 12 terms the partial sum is up to \(0.19935565\dots\) I would have gone further, but I was doing this on my phone and the OEIS list for the double factorials stops at that stage (and yes, I know I could have calculated more, but you get it). Even at that stage, the terms are still greater than \(0.001\), so convergence will be pretty slow, I think. At that rate I would anticipate needing another 10 terms or so just to get the answer correct to two significant digits.

Anyway, good luck to the participants in this fun event. You can probably guess where my sympathies lie, but all the entries so far have been pretty cool.