If is a prime integer, the numerator of the fraction is divisible by .
We wish to treat the given finite sum modulo , hence we can’t just add up fractions. We will have to consider each fraction as inverse of an integer modulo p. Observe that for , we have inverse of each element in the multiplicative group , i.e. there exist an such that .
For example, for , we have and .
Hence we have
for all .
Hence we have:
Thus we have:
The desired result follows by summation.
We can, in fact, prove that the above harmonic sum is divisible by , see section 7.8 of G. H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers for the proof.