Hello,  these are the numerous examples , not far from the original series, we are dealing here with 'multiples of zeros' ,  in other words, it resembles at the formula of Adamchik and Wagon, : from a specific form we can deduce an infinity of other like-formulas.  The current record of efficiency for my formula base 2 for Pi is the one from Huvent, https://fr.wikipedia.org/wiki/Formule_BBP see near the bottom of the page. And for the formula see formula (55) in http://mathworld.wolfram.com/PiFormulas.html and formula (48) with exponents = 65536^k Simon Plouffe Le 2018-04-06 à 00:29, françois mendzina essomba2 a écrit :

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...