Earlier this year I discussed a finite analogue of the harmonic sum. Today I wish to discuss a simple fact about finite harmonic sums.

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.