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We can use Bertrand's Conjecture ( that for any integer $n \not= 0$, there exists at least one prime number $$ with $n < p \leq 2n$ ) to demonstrate the following proposition:

  • $\forall n \in \mathbb{N}\setminus\left\{0\right\}$ , the rational number such that: $$ H_{2n} = \sum_{k = 1}^{2n}\frac{1}{k}\quad \mbox{is}\ \underline{never}\ \mbox{an integer} $$ What is the demonstration ?.
Felix Marin
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Alexis J
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    Note that, in that duplicate, the Bertrand based proof is described in the question itself. The posted solutions describe some non-Bertrand methods. – lulu May 27 '24 at 16:39

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Key idea: Write all the fractions in the sum with the common denominator $2n!$

Let $M_i$ denote the product $$1\cdot2\cdot3\cdot\ldots\cdot(i-2)\cdot(i-1)\cdot(i+1)\cdot(i+2)\cdot\ldots\cdot(2n-1)\cdot2n$$ i.e., $M_i = \dfrac{2n!}{i}$ for $1 \le i \le 2n$. Each $M_i$ is clearly an integer.

We can write $H_{2n}$ by the common denominator $2n!$ as $$H_{2n} = \dfrac{\sum\limits_{k=1}^{2n} M_i}{2n!}.$$

Using Bertrand's Postulate, Let $p$ be a prime such that $n < p \le 2n$. Notice that this implies $2p>2n$, so no other multiple of $p$ is among $\{ 1,2,\ldots,2n \}$.

Now, in the fraction $\dfrac{\sum\limits_{k=1}^{2n} M_i}{2n!}$, notice that:

  • The denominator $2n!$ is divisible by $p$.
  • For $i \neq p$, the term $M_i$ is divisible by $p$.
  • $M_p$ is not divisible by $p$.

Overall, in the fraction $\dfrac{\sum\limits_{k=1}^{2n} M_i}{2n!}$, the numerator is not divisible by $p$ but the denominator is divisible by $p$. Therefore, the fraction itself in not an integer. $\blacksquare$

Euclid
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