The essence of the series LC is that they exchange energy with each other. If you don't allow yourself to "look inside the box" then the series LC looks like a low impedance (zero, if ideal) at their resonant frequency. However, if you do look inside the box (the middle node where you pick off your choice to examine in your schematic) then you can see that this particular interior node is "dancing around" a lot, but where the voltages across the inductor and the capacitor are roughly equal and oppositely arranged to each other so that the voltage sum across the two appears to be close to zero.
What this kind of reminds me of is an old "thread & button spinner" toy I have played with from time to time. You can see it operate on this youtube video (I've specified it to start about where you can see it working well.) Only a little energy is being supplied by the slight pulling needed by the operator's hands every cycle, while the button rotates so fast you can hear the wind whistling by. The motion seen from both ends of the toy is modest. But the motion in the center is wild and fast!
Let's set this up in a simple LTspice schematic:
I'll explain the equations shown there and then get to the results.
The KCL is a little bit annoying, at first:
$$\begin{align*}
\frac1{L_1}\int V_{_\text{OUT}}\:\text{d}t + C_1\frac{\text{d}}{\text{d}t} V_{_\text{OUT}} &= \frac1{L_1}\int V_{_\text{GND}}\:\text{d}t
\\\\
\frac{V_{_\text{GND}}}{R_1} + \frac1{L_1}\int V_{_\text{GND}}\:\text{d}t &= \frac{V_{_\text{IN}}}{R_1} + \frac1{L_1}\int V_{_\text{OUT}}\:\text{d}t
\end{align*}$$
But if I take derivatives, solve and substitute a bit, and finally move things around I will wind up with:
$$\frac{\text{d}^3}{\text{d}t^3}V_{_\text{OUT}}+\frac{R}{L}\frac{\text{d}^2}{\text{d}t^2}V_{_\text{OUT}}+\frac1{LC}\frac{\text{d}}{\text{d}t}V_{_\text{OUT}}=\frac1{LC}\frac{\text{d}}{\text{d}t}V_{_\text{IN}}$$
By substitution of \$Z=V_{_\text{OUT}}^{\quad'}=\frac{\text{d}}{\text{d}t}V_{_\text{OUT}}\$ I find that the characteristic equation is \$Z^{''}+\frac{R}{L}Z^{'}+\frac1{LC}Z\$ with \$\omega_{_0}=\frac1{\sqrt{LC}}\$ and \$\zeta=\frac{R}2\sqrt{\frac{C}{L}}\$. Knowing that \$Q=\frac1{2\zeta}\$ I can find that if I know \$f_{_0}\$ and a desired \$Q\$, I can solve for the rest as shown in the above schematic.
I've assigned \$\tau_{_0}=\frac1{\omega_{_0}}\$. So \$L=\tau_{_0}\cdot Q\cdot R\$ and \$C=\frac{\tau_{_0}}{Q\,\cdot\, R}\$.
If I assign \$Q=1\$ then this is what I get:
Note that \$V_{_\text{GND}}\$ stays very close to ground. It doesn't move around that much. But now also note that \$V_{_\text{OUT}}\$ (the green trace, which is the capacitor voltage) and that \$V_{_\text{GND}}-V_{_\text{OUT}}\$ (the bluish trace, which is the inductor voltage) are moving almost exactly oppositely to each other and, broadly speaking, sum up close to zero all the time. Finally, note that all the magnitudes, including \$V_{_\text{IN}}\$, are the same.
Let's change to \$Q=2\$:
Note that \$V_{_\text{GND}}\$, while still close to ground, is swinging around a little bit more. Note that the peaks for the capacitor and inductor voltages are now twice as large as the input voltage peaks and still act in a way that their sum remains close to zero.
Let's try \$Q=5\$:
And now you can see that the peaks for the capacitor and inductor voltages are now five times as large as the input voltage peaks and yet still always sum to near zero. (\$V_{_\text{GND}}\$ has somewhat yet-larger swings.)
The key things to notice are that the voltage drop across the resistor remains very close to the source voltage and that the voltage drop across the LC pair remains very close to zero. What's happening "inside the box" is that energy is being transferred back and forth between the inductor and capacitor and that the rate at which the supply voltage itself is changing is just exactly what is needed so that the summed voltage across the inductor and capacitor remains close to zero throughout a cycle. If the source frequency were to be any higher, or any lower, then the timing of the exchange of energy between the inductor and capacitor would no longer match up so well and the effect would be lost.
