Category Archives: Finite harmonic sum

Nov10

Finite Sum & Divisibility – 2

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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 p is a prime integer, the numerator of the fraction 1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1} is divisible by p.

We wish to treat the given finite sum modulo p, 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 0<i<p, we have inverse of each element i in the multiplicative group \left(\mathbb{Z}/p\mathbb{Z}\right)^\times, i.e. there exist an i^{-1} such that i\cdot i^{-1}\equiv 1 \pmod p.

For example, for p=5, we have 1^{-1}=1, 2^{-1}=3, 3^{-1}=2 and 4^{-1}=4.

Hence we have
\displaystyle{i\cdot \frac{1}{i}\equiv 1 \pmod p ,\qquad (p-i)\cdot \frac{1}{p-i} \equiv 1 \pmod p }
for all 0<i<p.

Hence we have:
\displaystyle{i\left(\frac{1}{i}+\frac{1}{p-i}\right)\equiv i\cdot \frac{1}{i} - (p-i)\cdot \frac{1}{p-i} \equiv 0 \pmod p}

Thus we have:
\displaystyle{\frac{1}{i}+\frac{1}{p-i}\equiv 0 \pmod p}

The desired result follows by summation.

We can, in fact, prove that the above harmonic sum is divisible by p^2, see section 7.8 of G. H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers for the proof.

Jun25

Finite Sum & Divisibility

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I wish to discuss a small problem from The USSR Olympiad Problem Book (problem 59) about the finite sum of harmonic series. The problem asks us to prove that

\displaystyle{\sum_{k=2}^{n} \frac{1}{k}}  can never be an  integer for any value of n.

I myself couldn’t think much about how to prove such a statement. So by reading the solution, I realised that how a simple observation about parity leads to this conclusion.

Firstly, observe that among the natural numbers from 2 to n there is exactly one natural number which has the highest power of 2 as its divisor. Now, while summing up the reciprocals of these natural numbers we will get a fraction as the answer. In that fraction, the denominator will be an even number since it’s the least common multiple of all numbers from 2 to n. And the numerator will be an odd number since it’s the sum of (n-2) even numbers with one odd number (corresponding to the reciprocal of the number with the highest power of 2 as the factor). Since under no circumstances an even number can completely divide an odd number, denominator can’t be a factor of the numerator. Hence the fraction can’t be reduced to an integer and the sum can never be an integer.