Ragerin that resistor looks fine. Use the formula R(Resistance)=V(Voltage)/I(Current). Voltage in a series circuit, which this is, is calculated by taking your supply voltage (5v but you can only get multiples of 1.5v in standard batteries) and subtracting the forward voltage (voltage drop) of your LED (1.5v.) In this case you have three LEDs so take the 1.5v off three times. This leaves you with a calculation of R=5-(1.5*3)/0.015=34 Ohms Resistance. The 0.015 is the forward current in Amperes (15mA=0.015A).
Look for a resistor of at least 34 Ohms and you should get what you want. I have based these calculations of of the numbers you entered into the calculator. This should show a breakdown of what it did but I don't see why it gave you 39 Ohms. That must be a standard resistor. Those IR LEDs show a forward current of up to 100mA so for increased brightness you could increase to around 80mA due to the tolerance of resistors i.e. they are not perfect and the percentage of tolerance is how accurate they are so a 100 Ohm resistor with +/-5% could range from 95-105 Ohms. Always assume it will be the lower number (which could be why the calculator got 39 Ohms) but as long as you aim for a sensible proportional current less than the forward Voltage (IF) you should be fine.
That probably wasn't the best way to explain it so you can do a search on google for more information but thats the basics.
Hope it helps,
Boltz