# Divisibility properties of Harmonic numbers

Consider the $n$th harmonic number $$H_n = \sum_{i=1}^n \frac{1}{i}$$ It is a well known result that the $n$th Harmonic number is not an integer. Are there more generalized results for the Harmonic numbers? More specifically, if we collect the terms under a common denominator, $$H_n = \frac{m_n}{n!}$$ then the non-integer condition is equivalent to $n!\nmid m_n$. Can we prove stronger statements such as $(n-1)!\nmid m_n$? Can we generalize this statement to $(n-k)!$ for some $k$?

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Note the presence of the Stirling numbers. – anon Jun 27 '12 at 5:33
$m_n$ are listed at OEIS A000254 – Henry Jun 27 '12 at 6:19

## 1 Answer

The denominator of $H_n$ (in reduced form) will be divisible by every prime $p$ with $n/2\lt p\le n$ (and generally by lots of other primes as well). Bertrand guarantees there's one such prime, but in fact there are lots of them, the number goes to infinity as $n$ goes to infinity. It is certainly true that if $n$ is large enough (in terms of $k$) then $(n-k)!$ does not divide $m_n$.

By the way, the denominators of the harmonic numbers are tabulated at the Online Encyclopedia of Integer Sequences.

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