# Choosing $n$ and $r$ so that $n \choose r$ approximates $\pi$

I came across a curious formula, while trying out different numbers in $n \choose r$.

$${7.5 \choose 7} \approx \pi$$

The occurrence of $\pi$ with factorials has been discussed before, such as is in Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ ?

Using the gamma function or some other method, can we prove this approximate formula for $7.5 \choose 7$ ?

Also, is there any choice of $n$ and $r$ that yields $\pi$ exactly?

• Well, some value "out there" must be close to $\pi$, right? – barak manos Oct 13 '16 at 16:30
• I was excited when I first saw it, thinking it was exactly pi, but then I realized it was just close to it. So yeah, it might just be a coincidence. – Vermillion Oct 13 '16 at 16:37
• $$\binom{n+1/2}{m}$$ is a rational number for every $n,m\in\mathbb{N}$, hence it cannot equal $\pi$ exactly. – Jack D'Aurizio Oct 13 '16 at 17:25

Not a coincidence! $$\binom{7.5}{7}=\binom{7.5}{0.5} = \frac{\Gamma\left(\frac{17}{2}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma(8)}=\pi\cdot\left(\frac{16}{\pi\cdot 4^8}\binom{16}{8}\right)$$ hence $\binom{7.5}{7}\approx \pi$ is equivalent to $$\frac{16}{\pi\cdot 4^8}\binom{16}{8}\approx 1$$ that is a consequence of $$\frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}},\qquad \frac{16}{\pi\sqrt{8\pi}}\approx 1$$ so our approximation is essentially equivalent to $\pi^3\approx 32$, that follows from $$\frac{\pi^3}{32}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}$$ proved here. Actually, the last identity implies the tighter (and somewhat nicer) approximation $$\pi \approx 31^{1/3}.$$