Series Capacitors Equations
Given n capacitors in series (C1, C2, ..., Cn), the current through them is the same whereas the voltage divides depending on the capacitance of each cap.
The equivalent impedance of a series arrangement is the sum of the impedances, so:
$$ Z_{eq} = {-j · \left( { {1} \over {\omega · C_1} } + { {1} \over {\omega · C_2} } + ... + { {1} \over {\omega · C_n} }\right)} = -j · { {1} \over {\omega · C_{eq}} } \tag{1}$$
Taking out, the common terms:
$$ Z_{eq} = { {-j} \over {\omega} } · { \color{red}{ \left( { {1} \over {C_1} } + { {1} \over {C_2} } + ... + { {1} \over {C_n} }\right)} }= -{ {j} \over {\omega} } · \color{red}{ {1} \over {C_{eq}} } \tag{2}$$
By comparison, we get:
$$ C_{eq} = \left( { {1} \over {C_1} } + { {1} \over {C_2} } + ... + { {1} \over {C_n} }\right)^{-1} \tag{3}$$
If n=2, Eq. (3) gets simplified as:
$$ C_{eq} = { {C_1 · C_2} \over {C_1 + C_2} } \tag{4}$$