Calculating the dropper / series resistor for an LED
First of all, AAs are 1.5V, not 5V.

The equation is simple: V = I * R, where V is voltage, I is current, and R is resistance.

Start with your battery voltage and subtract the voltage drop of the LED itself.  This is the amount of voltage that your resistor must use up.  Then determine how much current you want to flow.  Probably between 0.001 and 0.020 amps (i.e., 1-20 milliamps, but the equations are for amps).

Now, just plug the numbers into Ohm's Law and go!

So, if you have a 9V battery, an LED that eats 2.1V, and you wanted 1 milliamp (0.001 amp), you would do: V = 9 - 2.1 = 6.9.  Ohm's Law:

V = I * R

6.9 = 0.001 * R 

R = 6.9 / 0.001

R = 6900.

If you have two LEDs (let's say they are both 2.1, and are in series with each other), then you have to subtract them both.  So V = 9 - 2.1 - 2.1 = 4.8.  Now do Ohm's Law again:

V = I * R

4.8 = 0.001 * R

R = 4.8 / 0.001

R = 4800

If you need more current, just change out the current in the equations.