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I was wondering whether, for each rational $q$, we may always write

$$q = \sum_{k=a}^b \frac 1k$$

For some positive integers $a \leq b$. I get the feeling that this is not true (although an immediate consequence of $\mathbb{R}^+$ being Archimedean is that the set of such $q$ is dense). I'm sure there is some slick proof using Bertrand's postulate (as is typical with these problems) but I'm not seeing it. This post is partly a reference request, as I'm sure this has been touched on before in some article, and would like to see it.

Descartes Before the Horse
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2 Answers2

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You cannot get $H_m-H_n=2$, or any integer $\ge2$.

In a sum $1/(n+1)+\cdots+1/m$ exactly one of the terms has minimal $2$-adic valuation. This can only be zero if one has only one term. Therefore there's a power of $2$ in the denominator, which cannot be $2^0$ save in the cases $H_{n+1}-H_n=1/(n+1)$ where $n+1$ is odd.

Angina Seng
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The proof given at ProofWiki and at Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer? for the fact that no harmonic number $H_n$ for $n\gt1$ is an integer also works for this case.

There is a unique denominator in $[a,b]$ that is divisible by the highest power of $2$. If $q$ is an integer, multiply through by one less factor of $2$ to turn all terms except one into integers (which is impossible). The only case to which this doesn’t apply is when the sum only runs over a single odd integer; and in this case the sum is also not an integer, except for $H_1=1$.

joriki
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