Yes, if you put the inductor and the peizo in parallel, then you have basically made a tank circuit. The set of calculations changes completely, the new set of calculations is Ω(total)=LC/L+C. The resonant frequency of the circuit is given by 1/(2π√LC).
I also got it a little wrong earlier (very tired, a nap helped). when the 2 are in series, their impedance is added, not canceled. As the frequency increases, the reactance of the inductor increases, but the reactance of the capacitor decreases so that is something that you need to consider. But the voltage drop across the 2 will remain proportional because of the voltage divider. You should also consider the resistor that you have placed in the circuit.