A bulb and a capacitor are connected in series to a source of alternating current. If its frequency increases keeping the voltage constant, the bulb will

The capacitive reactance of a circuit is given by:

${{\chi }_{c}}=\dfrac{1}{f}$

And the impedance of the circuit is given by:

$Z=\sqrt{R^2+\chi_c^2}$

When a bulb and a capacitor are connected in series to a source of alternating current. If the frequency of the source increases, then capacitive reactance decreases and thus the impedance of the circuit will also decrease.

If the impedance of the circuit decreases, the current in the circuit will increase and thus the bulb will glow brighter.