If you are only talking peak voltage output I would agree. But, since most (and I will accept being beaten up over this) AC measuring systems deal with "average" output over time:
the series resistance of the diodes is small
True
and the caps charge and discharge very quickly.
True and false. The caps charge quickly, but the output cap discharges slowly. Otherwise a charge pump wouldn't work. The discharge rate depends on the RC time constant of the output cap and the load resistance.
The charge is transferred near the edge of the input pulse.
. True and false. The charging current is available throughout the time that voltage, forward biasing a diode, is applied to the input. Once the voltage level on the input no longer forward biases the diode, the output cap discharges at an RC rate.
The duty cycle has little effect;
False. If I apply a short pulse (1% PWM) to the the pump, the cap is charged to peak, but has a long time to discharge before the next pulse. This would result in a very low "average" output voltage. If I have a 50% duty cycle, the cap is charged to "peak", but has less time to discharge, so the "average" voltage is higher. If I have a 99% duty cycle, the cap is charged to "peak", but almost no time to discharge. So the "average" voltage is essentially equal to the peak voltage.
And by you earlier logic the higher the frequency of the input, within limits, would result in shorter discharge times and therefore higher "average" voltage.
So...The devil's in the details.
But, we probably have the OP scratching both ends of their anatomy by now.
Ken