Seems about "twice as hard". (1/3)! == (2*2^(47/72)*((Pi*EllipticTheta[3, 0, E^(-((16*Pi)/Sqrt[3]))])/ (1/2^(3/16) + 1/(2^(7/16)*Sqrt[1 + Sqrt[3]]) + Sqrt[1 + Sqrt[3]]/(3 - 4*Sqrt[2] + 5*Sqrt[3])^(1/8)))^(2/3))/(3*3^(1/4)) Again just the terms through first order: 2 (1 + ----------------) Pi (16 Pi)/Sqrt[3] 47/72 E 2/3 2 2 (----------------------------------------------------------------) 1 1 Sqrt[1 + Sqrt[3]] ----- + ----------------------- + ------------------------------ 3/16 7/16 1/8 1 2 2 Sqrt[1 + Sqrt[3]] (3 - 4 Sqrt[2] + 5 Sqrt[3]) {(-)!, ------------------------------------------------------------------------------} 3 1/4 3 3 N[%, 51] {0.892979511569249211218564313658225881376229792652434, 0.892979511569249211218564313658225881376229792652429} --rwg