Yes you will need a current limiting resistor to stop the LED from frying.
to calculate:
Resistor(ohms) = (Battery voltage - LED voltage) / LED current (in amps)
example: if using a 12v battery and a 2.5v 20mA led:
(12-2.5)/ 0.02A = 475 ohms.... the closest resistor value in the E12 series will be 470 ohms.
The E12 series(DSE have them) have multiples of the following:
1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2
eg, 0.18(0R18), 1.8(1R8), 18, 180, 1800(1k8), 18k, 180k etc etc (R= Ohms, k = kiloOhms)
At 12v supply and 20mA, the resistor will be pretty close to it's power limit (1/4 Watt),
Power is voltage (dropped thru resistor) x current used, eg (12-2.5) * 0.02A = 0.2Watts
Two resistors in parallel will share the load, so two 1k 1/4W resistors in parallel will handle 0.5watts(1/4W * 2)and will have a resistance of 500 Ohms(1k / 2)
Hope that helps, just ask for more info if it's as clear as mud