From what I have previously posted:
J[4*n]=((4*n-3)/(4*n-2))* J[4*n-4] ; J[4*n+1]=((4*n-2)/(4*n-1))* J[4*n-3] ; J[4n+2]=((4*n-1)/(4*n))* J[4*n-2] ; J[4*n+3]=((4*n)/(4*n+1))* J[4*n-1] ; Verification of the result and sigma estimation J[4*n+3]=((4*n)/(4*n+1))* J[4*n-1] ; J[4*n+4]=((4*n+1)/(4*n+2))* J[4*n] ; By expressing the two preceding expressions according to J [3] and J [4], we will have: J[4*m+3]=J[3]*product(((4*n)/(4*n+1)),n=1..m) ; J[4*m+4]=J[4]*product(((4*n+1)/(4*n+2)),n=1..m) ; We have, on the same principles as for the sinus function: Limit(J[4*m+3],m=infinity) = Limit( J[4*m+4],m=infinity) ; Which corresponds to : J[3]*product(((4*n)/(4*n+1)),n=1..infinity)= J[4]*product(((4*n+1)/(4*n+2)),n=1..infinity) ; we obtain : J[4]/J[3]= product(((4*n)*(4*n+2)/(4*n+1)^2),n=1..infinity) ; we have : J[3]=1 et J[4]=(1/2)*J[0]=(1/4)*sigma Finally i find (1/4)*sigma = product(((4*n)*(4*n+2)/(4*n+1)^2),n=1..infinity) ; FME...