You can always put two LEDs in series (for 3.7V, not higher, see below for series explanation), they will each drop the voltage by ~1.6 volts and provide twice the light. The problem with one LED on 3.7 V is it will burn up because the voltage is too high, so the LED takes on too much current. When you put two in series, the first one consumes ~1.6v (varies by LED, look for threshold voltage), leaving 2.1v for the second LED (which is only slightly higher than the optimal voltage, but it will still outlive the clearo). Since the second LED consumes another ~1.6v, the first one doesn't receive the full 3.7 since the second is dropping the total voltage.
Each LED (in fact all diodes) require a threshold voltage in order to allow the current to pass through it, each diode (LED) will reduce the voltage by that threshold down the chain. For example, when I was a kid, I had a home stereo that put out 48 volts (at max audio output, that's was what the power pack had as the input voltage), I soldered a chain of 20 something LEDs in series, so they would light along with the bass. Didn't need a resistor because over the entire chain, they were all dropping the voltage together, now this meant I had to turn it up kind of loud before the voltage threshold was enough to jump across all LEDs (1.6 x # of LEDs).
To put them in series, just put the negative (-) lead from one LED to the positive (+) lead of the next LED. It's just like 2 AA batteries, plus to minus to add the voltage.
The second LED will be a little more dim than the first LED, but not by much. If you have a 5v battery, you could put 3 LEDs in series assuming they have a 1.6v threshold since 1.6 x 3 = 4.8v. However, as the battery gets weaker the voltage will drop below 4.8v and stop lighting them up, even though it could still operate the clearo fine for much longer. Of course, you could still do 2 in series, but would need a resistor.
To calculate the right resistor, you want to keep the amount of watts consistent with the wattage rating of the LED. So if you have a 500mW (0.5 watt) LED running at 3.7V (talking about a single LED only):
P = Power (watts)
E = Volts
I = Current (amps/milliamps)
R = Resistance (ohms)
Find the amount of amps necessary to generate 0.5 watts:
I = P / E - 0.5 (watts) / 3.7 = 0.13514 amps
Then find the resistance required to keep 3.7v flowing at 0.13514 amps:
R = E / I - 3.7 (volts) / 0.13514 (amps) = 27.31 ohms.
Now, keep in mind that the LED creates some resistance itself by using the power, so you can use a little lower ohms; I'm not exactly sure here, but maybe start with 10% less?
Google Ohm's Law Calculator for more formulas.
Hope this helps!
Edit: I forgot that I did have a resistor on my speaker LEDs. I did the same calculations above, but used 48V and the sum of the milliwatts (then converted to watts for the purpose of the formula) of all of the LEDs to derive the proper ohms to prevent them from frying.
I've got some parts ordered now and will have to try this when they get here.
