Ok, now I think I understand.
1) The DC component of your wave is the 0th harmonic, which will pass unmolested, as long as your filter is lowpass. So your average DC component will be the same independent of cutoff frequency. What you call ripple are the higher harmonics (1kHz and upper), which will become the more attenuated the lower your cutoff frequency is.
2) Look at Wenzel's Figure 2. The fundamental 1kHz harmonic is maximum for 50% duty cycle. That is the lowest "ripple" (non DC) frequency, so the most unaltered by the filter. It will dominate in the output.
3) Roghly speaking, the lower the cutoff and duty cycle, the lower the ripple, since you have lesser dominant terms (lower duty cycle) and all harmonics get more attenuated (lower cutoff). Your RC filter generally should reduce ripple voltage at a rate of about 3dB per octave of cutoff reduction below 1kHz.
For a 100% rigorous answer, you need the transfer function for your RC filter, generally Vout = Vin / (1 + jwRC), where w is the angular frequency and j the imaginary unit. Apply that frequency-dependent operator to the coefficients of the Fourier series of your variable duty-cycle wave (Wenzel, Fig. 1) and then compute the RMS norm of the result.