If x_(n+1) = (sqrt(1+x_n)-sqrt(1-x_n))/2
Then, using x_0 = 1/2, p_n = 2^(n+1) * 3 * x_n -> pi as n->inf
Additionally, p_n < pi.
If x_(n+1) = (sqrt(1+(x_n)*(x_n))-1)/x_n
Then, using x_0 = 1, p_n = 2^(n+2) * x_n -> pi as n->inf
Additionally, p_n > pi.
This seems pretty good but it's outclassed by the best pi algorithms.