Newton's method can rapidly sum the geometric series

Newton’s iteration for computing the reciprocal of a , y_{n+1} = y_n(2-ay_n), can be derived by applying Newton’s method to the function f(y)=1/y-a . Substituting a=1-x gives y_{n+1} = y_n(2-(1-x)y_n), an iteration for computing 1/(1-x) . Starting from y_0=1 , the first few iterates are \begin{aligned} y_1 &= 1 \cdot (1+x) = 1+x, \\ y_2 &= (1+x)(1+x^2) = 1+x+x^2+x^3, \\ y_3 &= (1+x+x^2+x^3)(1+x^4)…