March 14 is one of my least favorite days of the year. While most math teachers revel in Pi Day, I spend my 3/14s frustrated that we waste an opportunity to pay homage to one of the most important numbers in all of mathematics making endless nods to the fact that the name of a certain letter of the Greek alphabet, when mispronounced, is homophonous with a baked dessert. The entire premise of Pi Day is flawed anyway – it is based on $\pi$ being $3.14$, but that is only an approximation. So instead, I propose that we celebrate Approximation to Pi Day.

There are some really great approximations to $\pi$ out there. And there need to be: since the decimal expansion of $\pi$ goes on forever with no repeating pattern, if we didn’t have a decent estimate, we’d never be able to deal with $\pi$ at all. Schoolchildren all learn $22/7$ as their first estimate of $\pi$, and it’s not bad – since $\pi=3.1415926535\ldots$ and $\frac{22}{7}=3.1428571428\ldots$, this estimate at least gets the first two decimal places right. Other estimates are far more accurate. I wrote about The Wallis Product a while back, which can be used to estimate $\pi$ to any desired accuracy level.

One of the forgotten estimates of $\pi$ is that it is approximately $\sqrt{10}$. The estimate $\pi\approx\sqrt{10}$ is so accurate that in ancient times the two numbers were thought to be equal. Using modern calculators, we know that $\sqrt{10}=3.1622776601\ldots$, so it is a slight overestimate of $\pi$, but without those tools, it can be quite difficult to see that $\pi$ and $\sqrt{10}$ are distinct at all, let alone discern which is the larger. One reason it is hard to compare $\pi$ and $\sqrt{10}$ is that they tend to measure different things. We use $\pi$ to measure areas of lengths of curves, but $\sqrt{10}$, if anything, is usually the length of a side of a square. It would be helpful if we ever saw a $\pi^2$ in a formula somewhere so that we could just compare it to $10$. Luckily, $\pi^2$ does show up in a couple of places. For example, the volume of a $4$-dimensional hypersphere with radius $r$ is $\dfrac{\pi^2}{2}r^4$. But since $4$-dimensional space is hard to wrap our heads around, it might be easier to use a fact about infinite series: $$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots =\dfrac{\pi^2}{6}.$$ On the left side, we add up the reciprocals of the squares of all the positive integers, and after adding all of the infinitely many terms, we simply end up with $\pi^2/6$! If we want to know how $\pi$ and $\sqrt{10}$ compare, we can instead try to figure out how $1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots$ compares to $\dfrac{10}{6}$, or $\dfrac{5}{3}=1.66\overline{6}$.

Infinite sums are notoriously hard to evaluate exactly, so oftentimes, we rely only on estimates of them. With this particular sum, we can observe that

$$\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots < \dfrac{1}{6\cdot 5}+\dfrac{1}{7\cdot 6}+\dfrac{1}{8\cdot 7}+\dfrac{1}{9\cdot 8}+\cdots.$$

Each term of the sum on the right can be split into a difference of two terms:

$$\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{8^2}+\cdots < \left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{7}-\dfrac{1}{6}\right)+\left(\dfrac{1}{8}-\dfrac{1}{7}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\cdots.$$

When we drop the parentheses on the right side, everything cancels except the $\dfrac{1}{5}$, so we can say with confidence that $\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\cdots<\dfrac{1}{5}$. Therefore,

$$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\cdots< 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{5} =\dfrac{5989}{3600}=1.6636\overline{1}.$$

So, the infinite sum, the one equal to $\dfrac{\pi^2}{6}$, ends up being less than $\dfrac{5}{3}$, so $\pi^2<10$, and thus $\pi<\sqrt{10}$.