a BBP formula, I was surprised with the first identity by taking a = 0 and b = 14 to find: Pi=4*(sum(((-1)^(n+1)*cos((Pi*(2*n-1))/4))/((2*n-1)*2^((2*n-1)/2)),n=1..infinity))+2*sum(((-1)^(n+1)*sin((Pi*n)/2))/(n*2^n),n=1..infinity); who becomes : Pi =sum((-1/(8*n-1)-2/(8*n-2)-2/(8*n-3)+4/(8*n-5)+8/(8*n-6)+8/(8*n-7))*2^(2-4*n),n=1..infinity); Le mardi 3 avril 2018 à 21:53:44 UTC+1, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit : Hello For some values of a and b, especially values taken between 0 and 1 with a greater than b First identity ; theta=sum(((-1)^(n+1)*cos(a*theta)^(1-2*n)*sin(b*theta)^(2*n-1)*cos((b+a)*(2*n-1)*theta))/(2*n-1),n,1,inf)/b+sum(((-1)^(n+1)*sin(b*theta)^(2*n)*sin(2*(b+a)*n*theta))/(n*cos(a*theta)^(2*n)),n,1,inf)/(2*b); second identity; Pi/2-a*theta=sum((cos(a*theta)^n*sin((b+a)*n*theta))/(n*cos(b*theta)^n),n=1..infinity); With the second indentity, we can deduce many formulas: Pi/6=sum(sin((7*Pi*n)/12)/(n*2^(n/2)),n=1..infinity); Pi/6=sum(sin((2*Pi*n)/3)/n,n=1..infinity); Pi/6=sum(sin((8*Pi*n)/15)/(cos(Pi/5)^n*n*2^n),n=1..infinity); Pi/4=sum((2^(n/2)*sin((5*Pi*n)/12))/(n*3^(n/2)),n=1..infinity) ; We find the Madhava-Gregory-Leibniz formula, with a=1/4 and b=1/4 ; Pi/4=sum(sin((Pi*n)/2)/n,n=1..infinity); FME...