THD calculated with amplitude instead of RMS
Your colleague is correct.
$$
\sqrt{\sum_{n=2}^{\infty} V_{nrms}^2}=\sqrt{\sum_{n=2}^\infty \frac{1}{2}V_{npeak}^2}=\sqrt{\frac{1}{2}\sum_{n=2}^{\infty}V_{npeak}^2}
$$
As long as each \$V_{rms}\$ is referring to a sine wave.