Wallis' product exploration

[Allen]

Using identity 5.36 ( p. 186 of Concrete Mathematics )

\( \binom{n-\frac{1}{2}} {n} = \binom{2n} {n} ÷ 2^{2n} \)

I think we can also reduce the limit to:
\( \pi = \lim\limits_{n\to\infty} \frac {1} {n+\frac {1}{2}} ÷ \binom{n-\frac{1}{2}} {n}^{2} \)

(I could have messed something up there..)