Calculating harmonic distortion
THD=\$\sqrt{\frac{(V_2^2+...+V_n^2)}{V_1}}\$


The THD formula you have looks wrong to me so...

The RMS of all harmonics \$\sqrt{V_2^2 + V_3^2 + ... V_N^2}\$

Then this is divided by V1: -

THD = \$\dfrac{\sqrt{V_2^2 + V_3^2 + ... V_N^2}}{V_1}\$

This now makes sense because it's the ratio of two voltages - you had V1 within the sq root area and this is wrong.

So, to answer your question, yes the 3rd harmonic distortion is \$\dfrac{V_3}{V_1}\$