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The Title says is all tbh. I was trying to prove that Harmonic numbers can never be integers (except $1$) Pretty sure there exist better methods but whatever.

PS. Please don't use Bertand's Postulate

Martin Sleziak
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Anvit
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    The question is unclear. Take $p=2$ and $n=2$ so $2|2!$ but clearly $4$ does not. – Yanko May 27 '18 at 10:41
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    that theorem is called Bertrand's Postulate and is more-or-less equivalent to your statement. – Angina Seng May 27 '18 at 10:41
  • @Yanko I need to prove that this always happens – Anvit May 27 '18 at 10:44
  • @LordSharktheUnknown Wait, is it a theorem or postulate? – Anvit May 27 '18 at 10:46
  • It's a theorem, but it's known as a postulate because it was a postulate for a number of years before it was proved – Maxime Ramzi May 27 '18 at 10:46
  • Bertrand's Postulate was first showed by Tchebycheff then Erdös and Ramanujan had fun with it too. – Pagode May 27 '18 at 10:47
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    Just to say, the proof of Bertrand's Postulate is vastly more difficult than the proof that the harmonic numbers are never integers. See this question for elementary proofs of the latter claim. – lulu May 27 '18 at 10:53
  • Any prime, $p$, will satisfy the criteria such that $p\le n\le2p-1$, because since $p$ is prime, $n!$ will only be divisible by it when $n \ge p$. And $p^k$ will only be divisible by $n!$ when $n \ge kp$. – Badr B May 27 '18 at 11:04

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There can be no easy proof of your claim without Bertrand's postulate.

Indeed, assuming your statement to be true, here's a proof of Bertrand's postulate (which, as noted in the comments, is a theorem)

Let $p$ be a prime number and let $q$ be a prime number dividing $(2p)!$ such that $q^2$ does not divide it. $q\leq p$ is impossible because otherwise $2q\leq 2p$ and so $q^2 \mid (2p)!$.

Thus $q>p$. But also $q<2p$ otherwise $q$ would not divide $(2p)!$. Hence there is a prime $q\in ]p, 2p[$.

Now proceed by induction on $n$: there is clearly a prime between $2$ and $4$ ($3$ is prime).

Assume the statement is known for $n$ and we now want to show that there is a prime strictly between $n+1$ and $2(n+1)$.

Well there is a prime strictly between $n$ and $2n$ so the only way it could not be strictly between $(n+1)$ and $2(n+1)$ is if it's $n+1$. But if $n+1$ is prime, we know from what came above that there is a prime strictly between $n+1$ and $2(n+1)$, which is what we wanted to prove.

What this means is that there's "essentially" no way to prove this without Bertrand's postulate (or at least the proof is at least as hard as Bertrand's postulate's proof).

Now using Bertrand's postulate there's an easy proof of your statement: let $n\geq 2$ be any integer, and let $p$ be a prime strictly between $n$ and $2n$. Then clearly $p\mid (2n)!$ but $p^2\nmid (2n)!$.

However there are easier proofs of the claim you mention,as noted in the comments. Here's a link to some of them.

Maxime Ramzi
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