TITLE: Proof $\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}$ is not an integer for integer $n>1$
QUESTION [1 upvotes]: I found a way to prove this using Chebychev's theorem, are there ways to solve it without relying on this?

REPLY [0 votes]: 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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