Hi Greg,
Great post; good question.
First up, the voltage on the supply is so much higher than the voltage across the LED that the latter can be all but ignored when calculating the current. This current is essentially determined by the resistor alone.
So, if you want your LED to flow 5mA, and you have a 36V supply, you need a resistor around 36/5 kilohms, which is around 6.8K nearest preferred value. In truth, the LED will drop about two volts (1.6V for red, 2V for green, and around 2.2V for blue), and this is subtracted from the rail voltage to give the voltage across the resistor, which is actually 34V.
However, the rail voltage normally runs a volt or so over 36V, so the voltage across the LED is almost absorbed anyway.
Secondly, I cannot see any reason in physics why a bleeder would improve or damage the sound. However, it will slowly reduce the capacitor charge after switch off, and this is no bad thing. It won't take it to zero volts because the LED will stop passing current when it's trigger voltage is no longer met, as you surmised, but two volts across the caps won't do any harm and will hardly raise a sweat if you short out the caps with a screwdriver.
Lastly, if you want to discharge both rail caps, you'll need a bleeder on both. So, use a bleeder with an LED on say the positive rail, and a bleeder alone (of the same value) on the negative rail. I'd say something around 6K8 half watt is adequate; it will dissipate 190mW so this is quite adequately rated.
Cheers,
Hugh
