Wallis' product exploration

[Allen]

Nice, Thank you for checking in CAS!

of course at infinite limits the extra +1/2 is irrelevant, and since \( \binom{n-\frac{1}{2}} {n}^{2} = \binom{-1/2} {n}^{2} \), we can remove some more symbols.


as we can see on a discrete limit Wolfram alpha (or with continuous n)

\( \pi = \lim\limits_{n\to\infty} \frac {1} {n \binom{-\frac{1}{2}} {n}^{2}} \)

or circumference \(c \) is

\( c = \lim\limits_{n\to\infty} 2 \binom{-\frac{1}{2}} {n}^{-2} \)