Computing the voltage across a parallel LC component at resonance?
At resonance, the LC tank circuit behaves like an open circuit. 

Therefore, there is no current flowing through it. And therefore there is no current through the resistor.

Since there is no current through the resistor, there is no voltage across it. 

Therefore the full voltage of the voltage source appears across the LC tank.

To show specifically where your analysis went wrong, you said,


  $$V_{R} = ... = V \times \frac{R}{R - j(\frac{1}{0})}$$
  
  This results in an undefined solution though.


While this is mathematically undefined, you have a value going to infinity in the denominator of your right hand side expression. This means the total quantity goes to zero, so you have 

$$V_R = 0.$$

From there you can reach the conclusion I presented above.