DreamTeam wrote: Prove that for any $ n\in \mathbb N$ the number $ 1 + \dfrac{1}{3} + \dfrac{1}{5} + \ldots + \dfrac{1}{2n + 1}$ is not an integer.

Let $ m$ be the greatest integer such that $ 3^m\le n$. Hence all other numbers are of the form $ 3^k\cdot l$, where $ k<m$ and $ l$ is not divisible by 3. So if we take the lowest common denominator, the denominator is of the form $ 3^m\cdot j$; and all numerators, except the $ \frac1m$, are divisible by 3. So if we sum up, the numerator is not divisible by 3 and the denominator is divisible by 3, and hence the value is not an integer.

How about using Betrand's postulate? For any $n \geq 1$, there is a prime $p$ such that $n+1 < p < 2n+2$. From the bounds, it is clear that $p$ is odd; hence the term $\frac{1}{p}$ appears somewhere in the required sum. If we put all the terms under a common denominator: $1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n+1} = \frac{3 \dots (2n+1) + 1 \cdot 5 \dots (2n+1) + \dots }{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n+1)}$, Then all but one of the terms in the numerator are divisible by $p$. This means that the numerator is NOT divisible by $p$ while the denominator is, and therefore it cannot be an integer.

edit: this does not work since the 2-adic valuations are not necessarily distinct, I'll maybe fix this later

Remark: for the $n=2k+1$ it is trivial, because $1+\frac{1}{3}+\frac{1}{5}+\ldots +\frac{1}{2n+1}=\frac{2a}{2b+1}$