Behaviour of a passive filter
It can be helpful to look at this problem using voltage division.

Lets look at one of the simplest cases: an inductor and a capacitor in a voltage divider.







simulate this circuit &ndash; Schematic created using CircuitLab

Lets forget about the actual values of the inductor and capacitor. Lets assume that we are looking at a particular frequency and we have already converted the inductance and capacitance values into positive and negative reactances.

According the the rule of voltage division,

\$\frac{V_{out}}{V_{in}} = \frac{Z_1}{Z_1 + Z_2}\$

In this case,

\$Z_1 = \frac{1}{j2\pi fC} = -j50Ω\$

\$Z_2 = j2\pi fL = j40Ω\$

Again: I'm going to worry only about the reactances. I'm going to ignore the values of frequency, inductance, and capacitance in this example.

We can insert our values of impedance into the voltage divider equation like so:

\$\frac{V_{out}}{V_{in}} = \frac{-j50Ω}{-j50Ω + j40Ω} = \frac{-j50Ω}{-j10Ω} = 5\$

So this circuit has a voltage gain of 5 at the particular frequency we are concerned with.

If the inductive reactance and the capacitive reactance get closer and closer to being the same magnitude, then the voltage gain of the circuit will increase without bound. This is seen in the denominator of the voltage division equation: it tends toward zero when the reactances are made closer and closer to being equal and opposite.

In real life, the voltage gain is limited by resistances in the inductor and capacitor. To illustrate this, imagine what would happen to the voltage division equation if the reactances were +j49 and -j50 ohms. The denominator would be much smaller, making the voltage gain much higher.

p.s. You can do the same thing with positive and negative resistances.  ;)

Cheers