ac phasor voltage between capacitor and inductor higher than source voltage?
Somehow, the voltage at this location is higher than the source
  voltage. how could this be?


If you approach resonance in a series tuned circuit, the capacitive and inductive reactances are cancelling each other out. At resonance XL and XC together produce 0 ohms so, the current that flows is due only to V/R.

Clearly, at all other frequencies (higher or lower than resonance) the current is lower.

At resonance this high current flows through both L and C and you get voltage magnification - more current flows through C (due to L being present) than would flow through C if it were connected directly across the AC supply voltage.

Think of a mechanical equivalent. 



Imagine you had a mass of 1 kg suspended on a spring. You held one end of the spring (in the picture above that would be the solid line at the top) and gently moved that point up and down. If you did this much too quickly the mass would barely move (due to inertia). If you did it too slowly the mass would "follow" your hand movement.

If you did it at the goldilocks frequency the mass would start to move up and down and would progressively move up and down with a larger amplitude until it hits your hand or the spring breaks. 

You would find that you need very little hand movement (at resonance) to produce a large undulation in the mass.