I think the Equasion for caculatic LED resistors is:
(V {Batery} - V{LED}) / I {LED} = R
V {Batery} is the batery voletge
V{LED} is the LEDs forward voltege
I {LED} is the LEDs nomal curent (in amps,NOT miliamps)
R is the resistance reqierd (in Ohms)
For the "I {LED}" use a litle les the the maximum cntinues curent of the LED or the LED might overheat (burning out the LED) on long operatin
you should consider the led as a current operated device. Then decide how much current you want. For a normal red led it is somewhere around 20mA. The forward voltage on a normal red led is 2.1V, so:
(9V - 2.1V)/20mA = 345 ohms
In practice, you could use 330, 350, etc, they are all close enough. You can check the dimensionality too to make sure you are doing the right thing.
Yes, the lower the current the longer the battery life. It all depends on how bright you want the LED, the more current the brighter it is - so by reducing the current you reduce the brightness accordingly. You need to try higher value resistors until it's just bright enough for what you want.