Dear Pr Bill, Finally, the identity seems true for all Pi/4 == Sum[(Cosh[(1/2 + n)*y]*(2*n)!*((-Sqrt[2] + 2*Cosh[y])*Sech[2*y])^(1/2 + n))/ (2^(4*n)*((1 + 2*n)*n!^2)), {n, 0,∞}] Thank you... Le dimanche 18 août 2019 à 00:54:04 UTC+1, Bill Gosper <billgosper@gmail.com> a écrit : On Sat, Aug 17, 2019 at 2:21 PM françois mendzina essomba2 <m_essob@yahoo.fr> wrote: Hello, This identity seems true for some but not all?? values of x. But we can delete x. Pi/4=sum(((2*n)!*(cosh(5/(x/sqrt(x^17+1)+1))/(cosh(5/(x/sqrt(x^17+1)+1))^2-1/2)-1/(sqrt(2)*(cosh(5/(x/sqrt(x^17+1)+1))^2-1/2)))^((2*n+1)/2)*cosh(((5/2)*(2*n+1))/(x/sqrt(x^17+1)+1)))/((2*n+1)*2^(4*n)*n!^2),n=0..inf); best ragard Dr. E: You can change variables and get the less exotic assertion Pi/4 == Sum[(Cos[(1/2 + n)*y]*(2*n)!*((-Sqrt[2] + 2*Cos[y])*Sec[2*y])^(1/2 + n))/ (2^(4*n)*((1 + 2*n)*n!^2)), {n, 0,∞}] which I can neither prove nor refute.—Bill Gosper