Re: [math-fun] For lovers of Pi

another of this type sum((2^(-4*n-1)*(2*n)!*(2*cosh(2*x)+sqrt(3))^((-2*n-1)/2)*cosh((2*n+1)*x))/((2*n+1)*n!^2),n=0..infinity)=Pi/12 Le lundi 24 septembre 2018 à 18:45:10 UTC+1, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit : Hello, For all x, sum((2*n)!/(2^(3*n+1/2))/n!^2/(2*n+1)*(1/cosh(2*x)^(1/2))^(2*n+1)*cosh((2*n+1)*x),n = 0 .. infinity) = 1/4*Pi FME...