Yet another π formula

[Albert Chan]

Σ((-1)^k/(2k+1), k=0..inf) = atan(1) = pi/4
Σ((-1)^k/(2k+1)/(2k+3), k=0..inf) = (pi/4 - 0.5/(1)) / 1 = pi/4 - 1/2
Σ((-1)^k/(2k+1)/(2k+3)/(2k+5), k=0..inf) = ((pi/4 - 1/2) - 0.5/(1*3)) / 2 = pi/8 - 1/3
...

With this pattern, we can build formula for left hand side, S = pi / c1 - c2

lua> N = 40 -- number of consecutive odd numbers for LHS denominator
lua> c1, c2, t = 4, 0, 1 -- start from atan(1) = pi/4
lua> for i=2,N do c1=c1*(i-1); c2=(c2+0.5/t)/(i-1); t=t*(2*i-1) end

pi = c1*c2 + c1*S

lua> c1*c2 -- close enough to pi, we only need tiny correction.
3.141592653588783
lua> c1*c2 + c1 * (1-1/(2*N+1))/t -- 2 terms for S
3.141592653589793