The general form of this answer is correct, +1. However, you shouldn't be using the typical forward voltage drop, but the worst case (highest) for the current you want. The resulting calculation that you show than results in the minimum resistance to not exceed the maximum current rating of the LED.

Olin Lathrop‭ Wouldn't the worst case be the minimum voltage drop, not the maximum? The part didn't list a minimum so I used the typical value. It's better that a LED shines ever so slightly dimmer than supplying it with too high current. Also as I'm sure you know, In practice every modern LED out there can do with significantly lower current than 20mA which is the value for which they guarantee a particular luminous intensity. So if one uses bright modern LEDs with some >300mcd @ 20mA, they can be supplied with just a few mA and still shine like the sun.

Btw depending on what battery chemistry this is, then 9V fully charged might mean 10-11V. And down to somewhere around 7V when low. Obviously it isn't a good a idea to drive a LED directly from a raw battery voltage.

Oops. Yes, I meant to say "minimum".