Is there an elementary proof that n∑k=11k\sum \limits_{k=1}^n \frac1k is never an integer?

If n>1 is an integer, then nk=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.


Hint   Since there is a unique denominator 2K having maximal power of 2, upon multiplying all terms through by 2K1 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+232m+25+13+27MM

The prior sum has all odd denominators so reduces to a fraction with odd denominator d|357.

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.

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

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