If n>1 is an integer, then n∑k=11k is not an integer.

If you know Bertrand’s Postulate, then you know there must be a prime p between n/2 and n, so 1p appears in the sum, but 12p does not. Aside from 1p, every other term 1k has k divisible only by primes smaller than p. We can combine all those terms to get ∑nk=11k=1p+ab, where b is not divisible by p. If this were an integer, then (multiplying by b) bp+a would also be an integer, which it isn’t since b isn’t divisible by p.

Does anybody know an elementary proof of this which doesn’t rely on Bertrand’s Postulate? For a while, I was convinced I’d seen one, but now I’m starting to suspect whatever argument I saw was wrong.

**Answer**

**Hint** Since there is a unique denominator 2K having maximal power of 2, upon multiplying all terms through by 2K−1 one deduces the contradiction that 1/2=c/d with d odd, e.g.

m= 1+12+13+14+15+16+17⇒ 2m= 2+ 1+23+12+25+13+27MM⇒ −12 = 2+ 1+23−2m+25+13+27MM

The prior sum has all odd denominators so reduces to a fraction with odd denominator d|3⋅5⋅7.

**Note** I purposely avoided any use of valuation theory because Anton requested an “elementary” solution. The above proof can easily be made comprehensible to a high-school student.

**Attribution***Source : Link , Question Author : Anton Geraschenko , Answer Author : Bill Dubuque*