You are confusing the Q of the inductance with the Q of the circuit. In a series LCR circuit, the Q of the inductor is the ratio of the voltage across the inductor to the voltage across the circuit. As you say, this is also the ratio of the inductive reactance to the resistance. In a circuit with series R & L in parallel with C, this relation also holds.

In a parallel LCR circuit, the resistance is no longer a property of the inductor itself which is why you get a different definition for Q. The Q in this case is more a property of the circuit than the inductor. You cannot convert one circuit to the other although you could find two circuits with identical Q.

The Laplace transforms of the impedance for (r + L) // C is ...

$$\frac{\frac{r}{LC}+\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{r}{L}+\frac{1}{LC}}$$

for which Q=\$\frac{1}{r}\sqrt{\frac{L}{C}}\$ (at resonance)

The Laplace transform of the impedance for R // L // C is ...

$$\frac{\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{1}{RC}+\frac{1}{LC}}$$

for which Q=\$R\sqrt{\frac{C}{L}}\$ (at resonance)

You actually need just over 2 kHz sampling rate to sample 1 kHz sine waves properly. It's $$ f_N < f_S / 2 $$ not $$ f_N \le f_S / 2 $$

P.S. If you took your signal into complex space, where a sinusoid is of the form $$v(t) = Ae^{j(2 \pi f t - \theta)} = A(\cos(2 \pi f t - \theta) + j \sin(2 \pi f t - \theta))$$ where t is time, A is amplitude, f is frequency, and θ is phase offset, $$ f_N = f_S / 2 $$ is the point where the frequency "folds over", i.e. you cannot distinguish f from -f. Further increases in frequency will appear, after sampling, to have the sampling frequency subtracted from them, in the case of a pure sinusoid.

Non-Sinusoids

For the case of a square wave at 1 kHz with a duty cycle less than or equal to 10% which is sampled at 10 kHz, you are misunderstanding the input.

First you would need to decompose your waveform into a Fourier series to figure out what the amplitudes of the component harmonics are. You will probably be surprised that the harmonics for this signal are quite large past 5 kHz! (The rule of thumb of third harmonic being 1/3 as strong as the fundamental, and 5th being 1/5 of fundamental, only applies to 50% duty cycle square waves.)

The rule of thumb for a communications signal is that your complex bandwidth is the same as the inverse of the time of your smallest pulse, so in this case you're looking at a 10 kHz bandwidth minimum (-5 kHz to 5 kHz) for a 10% duty cycle with the fundamental at 1 kHz (i.e. 10 kbps).

So what will ruin you is that these strong higher-order harmonics will fold over and interfere (constructively or destructively) with your in-band harmonics, so it's perfectly expected that you might not get a good sampling because so much information is outside the Nyquist band.