This is in reference to an answer I gave to this question. I am curious to know if my intuition is correct. Given $n,m\in\mathbb{N}$, with $n>m$, is it possible for $\exp(2\pi in\sum\limits_{j=1}^n\frac{1}{j})=\exp(2\pi im\sum\limits_{j=1}^m\frac{1}{j})$ to hold, other than the solution where $n=2$ and $m=1$? This can be written equivalently as: are there any other solutions $n,m\in\mathbb{N}$, $k\in\mathbb{Z}$ with $n>m$ other than $n=2$, $m=1$ to the equation: $$\sum_{j=1}^m\frac{n-m}{j}+\sum_{j=m+1}^n\frac{n}{j}=k$$ I suspect that the answer is 'no,' and there may be a proof involving residues modulo $j$, though I would be interested in seeing any proof.
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$\sum 1/j$ is diverging so dont thinK theres any answer – Archis Welankar Mar 24 '16 at 03:45
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@ArchisWelankar There aren't any limits involved here. Notice these are the partial sums. – Ben Sheller Mar 24 '16 at 03:48
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There is at least one prime between $n$ and $2n$. That will appear in the denominator of exactly one term. So there are no solutions. Now that does not immediately work for $n$ odd, but Bertrand's Postulate is a weak result, so I suspect you could turn that into a rigorous proof with a little juggling. – almagest Mar 24 '16 at 11:00