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

Thank you Pr Bill for the exception...here is another identity, I noticed that it seems to converge on Pi / 2, Pi/2=sum((2^((1-6*n)/2)*(2*n)!*sech(2*x)^(n+1/2)*cosh((2*n+1)*x))/((2*n+1)*n!^2),n=0..infinity); this identity converges quickly for x in] -1,1 [ I suppose it also admits an exception Le mardi 20 août 2019 à 03:23:07 UTC+2, Bill Gosper <billgosper@gmail.com> a écrit : On Mon, Aug 19, 2019 at 1:17 PM françois mendzina essomba2 <m_essob@yahoo.fr> wrote: 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 : Except <odd> I π (!).  —Bill  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