Looking at the first infinite product of pi found by François Viet, I deduced an infinite product of atan(x) (2/sqrt(2/(17*sqrt(5))+2))*(2/sqrt(sqrt(2/(17*sqrt(5))+2)+2))*(2/sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(17*sqrt(5))+2)+2)+2)+2)+2)+2)) * ... =(17*sqrt(5)*atan(38))/38; (2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2)+2))* (2/sqrt(sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2)/5+2)+2)+2))*(2/sqrt(sqrt(sqrt(2)/5+2)+2))*(2/sqrt(sqrt(2)/5+2)) * ... =(5*sqrt(2)*atan(7))/7 ; And in a general way (2/sqrt(2/(sqrt(x^2+1))+2))*(2/sqrt(sqrt(2/(sqrt(x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2)+2))* ... = sqrt(1/x^2+1)*atan(x) ; An approximation of order 6 will be (2/sqrt(2/(sqrt(x^2+1))+2))*(2/sqrt(sqrt(2/(sqrt(x^2+1))+2)+2))*(2/sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2))*(2/sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2/(sqrt(x^2+1))+2)+2)+2)+2)+2)+2)) = sqrt(1/x^2+1)*atan(x);