LEDs have a forward operating voltage "Vf".This varies with current but is around 3V for both blue and white LEDs at typical operating currents.
If you place 2 LEDs in series they would require about 6 volts to provide a Vf of 3 V for each.
In this case you are fortunate that the supply voltage has been current limited enough not to destroy the LEDs. Applying voltage sources well above normal operating Vf will allow large currents to flow and will usually destroy the LED.
Proper practice is to provide a series resistor to drop the extra voltage and to limit current.
Voltage drop is Vsupply - Vf = Vr.
Resistor value= V/I = Vr / I_wanted.
Here if Vsupply = 5V, I = 10 mA, Vf = 3V.
R = (5-3)/0.010 = 200 Ohms (approximately).
Source current limitations will reduce actual current when supply is from eg processor pins.