Yeah, you're right. A bit of algebra reveals that your delta-V per MASS of fuel (which is a frame-independent quantity) goes as the (propellant specific chemical energy)^.5.
However, it is also true that if you calculate the delta-V per unit propellant ENERGY used, it decreases with the propellant specific energy.
Since you don't really care about how much energy you carry but only about how much fuel you need to use per delta-V, it's always better to have higher exhaust energy. (As long as you are using a chemical propellant, i.e. the mass carrying the energy and the reaction mass are the same.)
The statement in the article about "propulsive efficiency" is total bogus. First: the relative kinetic energy of the rocket and the exhaust IS a frame-dependent quantity, so it's meaningless. Second, any reasonable measure of efficiency, like the delta-V per fuel energy, reveals that the "optimal" value is an exhaust velocity of zero. (Optimal in the sense that it's more energy efficient, but only because the delta-V goes to zero slower than the energy. It's still a useless optimum, because it results in no delta-V at all...)
However, it is also true that if you calculate the delta-V per unit propellant ENERGY used, it decreases with the propellant specific energy.
Since you don't really care about how much energy you carry but only about how much fuel you need to use per delta-V, it's always better to have higher exhaust energy. (As long as you are using a chemical propellant, i.e. the mass carrying the energy and the reaction mass are the same.)
The statement in the article about "propulsive efficiency" is total bogus. First: the relative kinetic energy of the rocket and the exhaust IS a frame-dependent quantity, so it's meaningless. Second, any reasonable measure of efficiency, like the delta-V per fuel energy, reveals that the "optimal" value is an exhaust velocity of zero. (Optimal in the sense that it's more energy efficient, but only because the delta-V goes to zero slower than the energy. It's still a useless optimum, because it results in no delta-V at all...)