Robert Israel:
>Of course there's also
>pi = ((-1)^(n+1) 2 Zeta(2 n) (2 n)! /bernoulli(2 n))^(1/(2 n))/2
>so
>pi ~= ((-1)^(n+1) 2 (1 + 2^{-2n} + ...) (2 n)! /bernoulli(2 n))^(1/(2 n))/2
Apropos Zeta, what about a formula with Zeta(1/2)?
Ok, here is the full story:
http://www.luschny.de/math/euler/GammaQuotientsAsRadicals.html
Peter