I've been told that, generally speaking, rotor efficiency increase with rotor diameter. This is because the thrust generated by a helicopter rotor equals the mass of air moved times the delta V of the air moved, while the kinetic energy imparted to the air by the helicopter is proportional to the mass of the air times the square of the delta V. Therefore, since increasing the rotor area increases the mass of air moved, and since hovering requires a constant thrust force, increasing the rotor area will decrease the delta V of the air. And according to E = 1/2 m*v^2, doubling the mass and halving the delta V of the air will decrease the energy imparted to it.
From this, you can theorize that a helicopter with an infinitely large rotor diameter would require zero energy to remain hovering.
Now, assuming this is all correct, let's assume that the helicopter briefly accelerates upwards until it is moving vertically at 1 mph. Let's also assume that wind resistance is negligible. As it moves steadily upward, it will need the same force as when it was just hovering. And the above analysis suggests that the helicopter with the infinitely large rotor would not need to consume any energy to maintain the constant upward speed. But we know that a helicopter climbing into the sky is gaining potential energy, meaning that this conclusion of zero work done by the helicopter can't be correct.
So, what am I missing? Why does the analysis that suggests that neither a hovering helicopter nor a steadily rising helicopter with an infinitely large rotor need to consume any energy?