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I found a way to prove this using Chebychev's theorem, are there ways to solve it without relying on this?

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marked as duplicate by Thomas Andrews, Thomas, Emily, azimut, Jorge Fernández Hidalgo Jul 24 '13 at 14:33

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    $\begingroup$ See this. $\endgroup$ – David Mitra Jul 24 '13 at 13:37
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Hint: Pick the largest $m$ so that $2^m \leq n$.

Isolate $\frac{1}{2^m}$ and add all the other fractions. Then your sum will have the form

$$\frac{1}{2^m}+\frac{k}{l}$$

where $2^m \nmid l$.

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assume $\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}=A$ with A is integer. Note that between $2 \leq i \leq n$, there must be i such that $i = 2^k$, with largest k thus $2^k| LCM(2,3,4,....,n)$ then each side(LHS and RHS) times $LCM(2,3,4,....,n)$ Obviously, LHS is even Look at RHS, $LCM(2,3,4,....,n) \cdot (\frac{1}{2}+\frac{1} {3}+...\frac{1}{n})$. we get all of the number are even except $LCM(2,3,4,....,n) \cdot \frac{1}{i}$ is odd. So, RHS is odd. Contradiction.

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