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


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

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


where $2^m \nmid l$.


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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