To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $ R $. At resonance, the impedance $ Z $ of the series LCR circuit is equal to the resistance $ R $, and thus:

$$ Z = R $$

The amplitude of the current $ I_0 $ at resonance is given by Ohm's law:

$$ I_0 = \frac{V_0}{R} $$

where $ V_0 $ is the amplitude of the voltage supplied.

Now, if the resistance $ R $ is halved while keeping the voltage amplitude $ V_0 $, capacitance $ C $, and inductance $ L $ the same, the new resistance becomes:

$$ R_{new} = \frac{R}{2} $$

The new current amplitude $ I_0' $ at resonance is given by the modified Ohm's law:

$$ I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0 $$

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:

Option D: double