1. Condition at resonance in a series LCR circuit

At resonance, $X_L = X_C$ so the net reactance is zero.

Hence the impedance becomes purely resistive: $Z = R$

2. Current amplitude at resonance

If the applied voltage amplitude is $V$, then the current amplitude is $I = \frac{V}{Z} = \frac{V}{R}$

So, at resonance, $I_{\text{res}} \propto \frac{1}{R}$

3. Resistance is halved

Initially, $I_1 = \frac{V}{R}$

When resistance is halved, $R' = \frac{R}{2}$

Then the new current amplitude at resonance is $I_2 = \frac{V}{R/2} = \frac{2V}{R} = 2I_1$

4. Conclusion

The current amplitude becomes double.

Therefore, the correct option is: $\boxed{\text{D: double}}$