# A curious product formula

Fiddling with Mathematica seems to suggest the following:

$$\frac{(2^2)(4^2)(6^2)\cdots(2N^2)}{(1^2)(3^2)(5^2)\cdots(2N-1)^2}=N\pi+\frac{\pi}{4}+\frac{\pi}{32N}-\frac{\pi}{128N^2}+o(1/N^2).$$

Does anyone have an explanation or reference for this, or know how the series continues? The next term appears to be around $-\frac{\pi}{411.5N^3}\approx-0.00763$.

• How did you got such an approximation? – chubakueno Dec 6 '13 at 4:08
• I was looking up the Taylor coefficients for arcsin(x) and started playing with them. I squared everything because the series on the left looks nicer that way. – Rob Silversmith Dec 6 '13 at 16:36

## 2 Answers

Your first term is $r^2$, where $$r=\frac{2\cdot 4\cdots (2n)}{1\cdot 3\cdot 5\cdots (2n-1)}=\frac{(2\cdot 4\cdots (2n))^2}{(2n)!}=\frac{4^nn!n!}{(2n)!}=\frac{4^n}{{2n\choose n}}$$

Now, ${2n \choose n}$ is the central binomial coefficient, which is known to be approximately $\large \frac{4^n}{\sqrt{\pi n}}$ for large $n$.

Hence $r^2\approx \pi n$. You are teasing out more terms of the approximation (then squaring). You can get as many terms as you like by using terms of the Stirling series to approximate the factorials.

• PS. This is a pretty neat thing to discover on your own, so kudos Rob. – vadim123 Dec 5 '13 at 20:53

This is similar to the famous Wallis product, discovered by John Wallis three or four centuries ago, and it can be seen as a special case of Euler's infinite product for the sine function. $$\frac{\sin\pi x}{\pi x}=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)$$ Take $x=\frac12$ , and Wallis' identity follows.