Voltage output across purely capacitive voltage divider when PWM/Square wave signal is applied?
Let's assume there is some series resistance to stop the Universe from exploding.

When voltage is applied across capacitors in series, the voltage splits in inverse proportion to their capacitance. The reason for this is that in a series circuit the same current flows through both capacitors, putting the same amount of charge in each. Since the formula for charge on a capacitor is \$q = CV\$, a smaller capacitor will have a higher voltage for the same charge.

With some resistance in series the current will start at a value set by the total resistance and applied voltage, and the capacitor voltages will rise exponentially - just like it does with a single capacitor. If the resistances are internal to the capacitors then their terminal voltages will have an initial step due to voltage drop across the resistance, but once they have fully charged their voltages will settle down in inverse proportion to their capacitance.

A good example of a practical capacitive divider is the 10x probe on an oscilloscope. At DC and low frequencies the division ratio is set to 10:1 by a 9M&ohm; and 1M&ohm; resistor. However at high frequencies the cable capacitance changes the division ratio, distorting the waveform. To compensate for this a trimmer capacitor set to ~9 times less capacitance than the cable is wired across the 9M&ohm; resistor, restoring the 10:1 division ratio.