I simplified the symmetric identity in http://www.tweedledum.com/rwg/stanfordn3.pdf : a x z + ContinuedFractionK[a ((x + n)^2 + b) z^2, ((a - 1) (x + 1 + n) - (a + 1) y + 1) z, {n,0,∞]}] == a y z + ContinuedFractionK[a ((y + n)^2 + b) z^2, ((a - 1) (y + 1 + n) - (a + 1) x + 1) z, {n,0,∞]}] With {x -> 1, z -> z/Sqrt[a], b -> 0, a -> (2 + z^2 - 2 Sqrt[1 + z^2])/z^2, y -> 1/2} this gives z/ArcTan[z] == 1/2 (1 + Sqrt[1 + z^2]) + ContinuedFractionK[(1/2 + n)^2 z^2, 2 + 2 n + Sqrt[1 + z^2], {n, 0, ∞}] or ContinuedFractionK[1/4 (1 - 2 n)^2 (-1 + z^2), 2 n + z, {n, 1, ∞}] == -1/2 - z/2 + Sqrt[-1 + z^2]/ArcSec[z] And with ContinuedFractionK[c*n^2 + d*n + e, a*n + b, {n, 0, Infinity}] == (2*c*e* Hypergeometric2F1[1 + (2*e)/(d + Sqrt[d^2 - 4*c*e]), 1 + (d + Sqrt[d^2 - 4*c*e])/(2*c), (1/2)*(3 + d/c + (2*b*c - a*(c + d))/(c*Sqrt[a^2 + 4*c])), 1/2 - a/(2*Sqrt[a^2 + 4*c])])/((2*b*c - a*(c + d) + Sqrt[a^2 + 4*c]*(c + d))* Hypergeometric2F1[(2*e)/(d + Sqrt[d^2 - 4*c*e]), (d + Sqrt[d^2 - 4*c*e])/(2*c), (1/2)*(1 + d/c + (2*b*c - a*(c + d))/(c*Sqrt[a^2 + 4*c])), 1/2 - a/(2*Sqrt[a^2 + 4*c])]) we get an identity relating four 2F1s. --rwg And the insane, multiminute delays plaguing Sum and Product now also infest ContinuedFractionK. Since Mma knows so little about ContinuedFractionK anyway, just replace it with the undefined function cfk. (Anybody remember Charles F. Karney?)