If driving a LED with a series resistor, you need a voltage that is above the forward voltage of the LED at your desired operating current by a sufficient margin to account for LED variability.
Unfortunately component manufacturers often under-specify their products. In your case they give a graph for the typical forward voltage at different currents, but they only specify a maximum forward voltage at one particular current level and they don't specify a minimum forward voltage at all.
So lets try some designs. Lets say we want to aim for a 10mA forward current. That means a forward voltage of around 1.82V.
With one LED on a 5V supply, that gives us 3.18V for the resistor, and a desired resistor value of 318 ohms. Round that to the closest common value of 330 ohms.
This will be a stable but inefficient configuration, more voltage is dropped over the resistor than the LED, so variation in LED forward voltage will have little impact.
Now lets try two LEDs in series. Now 3.64V is taken by the LEDs, so that leaves only 1.36V for the resistors. Now we only need a 136 ohm resistor to limit the current.
This system will be more sensitive to variability in the LEDs and in the supply voltage, but it will probably still be ok in practice. At least for indicator LEDs there is a pretty wide range of currents that will result in acceptable brightness.
With three LEDs in series we have a problem. 1.82*3 = 5.46. So with LEDs that perform typically we won't achieve our goal of 10ma, even with no resistor at all! The graph becomes hard to read at that level but it looks to me like the current will maybe be around 1mA at that voltage. This is likely to be very variable with slight changes in supply voltage and temperature and between different LEDs of the same model.