A curious product formula

Question

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 $-\frac{\pi}{411.5N^3}\approx-0.00763$.

Answer 1

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.

Answer 2

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.