Can $n! $ divide $ n! + \frac{n!}{2} + \dots + \frac{n!}{n} $? 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.
 A: 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 $n < q < 2p$.
A: 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
A: 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? 
