Why do capacitors lose capacitance in series?
The answer to this comes from considering what is capacitance: it is the number of coulombs (C) of charge that we can store if we put a voltage (V) across the capacitor.

Effect 1: If we connect capacitors in series, we are making it harder to develop a voltage across the capacitors. For instance if we connect two capacitors in series to a 5V source, then each capacitor can only charge to about 2.5V.  According to this effect alone, the charge (and thus capacitance) should be the same: we connect two capacitors in series, each one charges to just half the voltage, but we have twice the capacity since there are two: so break even, right? Wrong!

Effect 2: The charges on the near plates of the two capacitors cancel each other. Only the outer-most plates carry charge. This effect cuts the storage in half.

Consider the following diagram. In the parallel branch on the right, we have a single capacitor which is charged. Now imagine that if we add another one in series, to form the branch on the left. Since the connection between the capacitors is conductive, bringing the two plates to the same potential, the ----- charges on the bottom plate of the top capacitor will annihilate the +++++ charges on the top plate of the bottom capacitor.

So effectively we just have two plates providing the charge storage. Yet, the voltage has been cut in half.



Another way to understand this is that the two plates being charged are farther apart.  In free space, if we move plates farther apart, the capacitance is reduced, because the field strength is reduced.  By connecting capacitors in series, we are virtually moving plates apart. Of course we can place the capacitors closer or farther on the circuit board, but we have now have two gaps instead of one between the top-most plate and the bottom-most plate. This reduces capacitance.