Take the harmonic number
$$\sum_{n=1}^m \frac{1}{n}=\frac{a}{b}$$
Define $a$ and $b$ as the actual integers produced by the summation - i.e., treat the numerator and denominator separately and do no clear like terms. Are $a$ and $b$ coprime?
It is clear that $b=m!$. But is $\frac{a}{m!}$ automatically in lowest terms 'straight out of the box'?
Instinct says yes, because there will always be a term in the numerator's sum that has a factor that is not shared with the other terms. But instinct is fallible. I've tried to read around, but I'm still not 100% clear; the assumption always seems to be that the fraction is in lowest terms.
Is my instinct correct? How would one prove it?