Can $n! $ divide $ n! + \frac{n!}{2} + \dots + \frac{n!}{n} $?

Question

How to disprove that $$n! \mid \left(n! + \frac{n!}{2} + \dots + \frac{n!}{n} \right)$$ This should not be true, since it would imply that there is some $n$ such that the $n$-th partial sum of the harmonic series reaches an integer.

Here is what I tried:

Since $n > 2 $ (Otherwise this would me trivial), then: $$ 2 \mid n! \\ 2 \mid \frac{n!}{2} $$ So that $n > 4 $ (Note that $n$ can't be $4$). Then, by the same reasoning: $$ 4 \mid n! \\ 4 \mid \frac{n!}{4} $$ So that $n > 6 $ But then I get stuck here since $ 6 \mid \frac{n!}{6} $ doesn't necessaily make $n$ larger than 6.

Answer 1

Let $p$ be the largest prime less than $n$. Then $$ \sum_{i = 1}^n \frac{n!}{i} \equiv \frac{n!}{p} \not\equiv 0\mod p $$ Since $n! \equiv 0\mod p$, it cannot divide that sum.

EDIT: The comments tell me I should explain why $i = 2p$ doesn't appear in the sum. Bertrand's postulate says there is a prime $q$ with $p < q < 2p$. Since $p$ is the largest prime less than $n$, we must have $n < q < 2p$.

Answer 2

For $n\gt 1$ Betrand's postulate shows there is at least one prime $p$ greater than $\frac{n}{2}$ and less than or equal to $n$

So $\left(n! + \frac{n!}{2} + \dots + \frac{n!}{n} \right)$ is the sum of $(n-1)$ multiples of $p$ and one non-multiple of $p$, so is not a multiple of $p$

But $n!$ is a multiple of $p$ so does not divide the sum

Answer 3

Here is a nice proof I gave on my own, of the fact that $H_n$ is not an integer for any $n$. Let $p$ be the greatest prime not exceeding $n$. Now, if you calculate $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ by long division, the denominator will be $n!$ and the numerator should be a sum of terms like $\frac{n!}{i}$, where $i$ runs from $1$ to $n$, i.e. $$H_n = \frac{\frac{n!}{1}+\frac{n!}{2}+...+\frac{n!}{n}}{n!}~.$$ Now, the denominator is divisible by $p$. In the numerator, all terms other than $\frac{n!}{p}$ are divisible by $p$, and $\frac{n!}{p}$ is of course not divisible by $p$. Hence, the numerator is not divisible by $p$. This proves that $H_n$ is not an integer.

Note: The proof is slightly incomplete. Can you point out where?