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

can never be an integer for any value of .

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 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 . And the numerator will be an odd number since it’s the sum of 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.

### Like this:

Like Loading...

*Related*

If I remember correclty, if we do not have the restriction that we need to choose consecutive reciprocals, then any integer can be represented by the sum of reciprocals. That is, N = sum (1/i) where {i} is some subset of the natural numbers.

LikeLiked by 1 person

Yes, you are right 🙂 Such fractions are called Egyptian fractions, see: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html

There are many nice solved and unsolved problems involving them, see the Wikipedia and Wolfram MathWorld pages.

LikeLiked by 1 person

Pingback: Finite Sum & Divisibility – 2 | Gaurish4Math

whoops, I meant ‘if n is odd’ not ‘if t is odd’

LikeLike

Wow! That’s a neat solution. I think I solved it in a different way, but it involves some heavy technology…

Suppose for contradiction that 1/2 + 1/3 + … + 1/n = k for some integer k.

Multiply by n!. This gives n!/2 + n!/3 + … + n!/n = n!k.

Bertrand’s postulate – it’s been proved by Ramanujan! 🙂 – states that for each positive integer t, there exists a prime p greater than t and at most 2t. If n is even, we take t = n/2. If t is odd, we take t = (n – 1)/2. These both tell us that there exists a prime p greater than n/2 but at most n. Thus in the equation n!/2 + n!/3 + … + n!/n = n!k, every term is divisible by p except n!/p. This cannot be divisible by p as none of the integers 2, 3, … n are divisible by p except p itself, since n is less than 2p.

Now we have a contradiction since the right hand side of the equation n!/2 + n!/3 + … + n!/n = n!k. is divisible by p, but the left hand side is not.

Does that work? I’m sure there must be a way to do a similar proof without using Bertrand’s postulate…finding a prime that divides every term but one…

LikeLiked by 1 person

Sure the solution is correct. Just to complete, here is Erdős’s elementary proof of the postulate: https://gaurish4math.files.wordpress.com/2014/12/erdos.pdf

Before going through the solution from the book, I tried solving it by the method of contradiction. But I failed to achieve that using some high-school level arguments. I didn’t even imagine that Bertrand’s Postulate would be useful. If you find a way of achieving contradiction which avoids Bertrand’s postulate, do let me know.

LikeLike