Electronic – Why is the AC coupling capacitor always 0.1uF or 0.01uF in high speed transceivers

I didn't read your whole question, which seemed to go out of its way to make a simple thing complicated. As I understand it, you have a capacitor, resistor, and LED all in series, and you want to know how things decay if the capacitor starts out initially charged up.

At first apporoximation, you can consider the LED a voltage source. That means the current will decay just as if the LED wasn't there and the cap was charged up to the LED voltage less than what it really was. This is now a simple R-C systems which follows a basic exponential decay with a time constant of RC, which it seems you already understand. The question of when the LED goes "off" then comes down to at what current you consider the brightness to be low enough to be off. This can vary a lot by the efficiency of the LED, ambient light level, and how obvious "on" is supposed to be. For example, if the cap is initially charged so that the initial current is 20 mA (a common maximum for LEDs) and you consider 1 mA the "off" level, then the on time will be the 95% decay time, which is 3.0 time constants.

As I said, this was the basic first approximation where the LED has a fixed voltage accross it. That will be largely true, but of course its voltage will drop with current somewhat. For practicle purposes, this is a small effect compared to the slop of deciding what current level "off" really is, unless that current is small, like less than a mA.

A common way to deal with this is to use a "DC servo" circuit to cancel out the DC component. Build a low-pass filter to isolate the DC and then invert it and add it to the original signal. The advantage is that you can use high values of resistance and relatively low values of capacitance in the low-pass filter.