# 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.

• Can you prove that $H_n\notin\mathbb{N}$? – Michael McGovern Jan 25 '18 at 1:08
• That would of course prove my question! But I would like to see if my attempt to prove that no harmonic number is an integer, has a future. – I.Padilla Jan 25 '18 at 1:08
• Brief sketch: If $2^k\le n<2^{k+1}$ then $2^k(1+\cdots+1/n)\equiv1\bmod 2$. – anon Jan 25 '18 at 1:11
• How about $\gcd(m,n)=m\rightarrow{n}\ndiv{m}$? – Michael McGovern Jan 25 '18 at 1:11

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 $n < q < 2p$.

• How did you get away without invoking Bertrand's postulate? – user491874 Jan 25 '18 at 1:20
• I just noticed that, in order for this $\frac{n!}{p} \equiv 0 \text{mod$p$}$ to be a contradiction, we would need BP! – I.Padilla Jan 25 '18 at 1:25
• @user8734617 Padilla, is right how do you know the power of $p$ in $n!$ is not higher than $1$ if not for Bertrand's postulate? – clark Jan 25 '18 at 1:52
• @user8734617 I may have assumed that was obvious. Perhaps I should have mentioned it. – eyeballfrog Jan 25 '18 at 2:09
• Yep, would be good IMHO, because "obvious for you" and "obvious for the OP" are two different things. – user491874 Jan 25 '18 at 2:18

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

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?

• Yep, at the point where you claim that the only term not divisible by $p$ is $\frac{n!}{p}$. If (contrary to Bertrand's postulate) there were no bigger primes between $p$ and $2p$, and if it happened that $2p\le n$, then you would have another term $\frac{n!}{2p}$ not divisible by $p$. – user491874 Jan 25 '18 at 1:19
• Yes, you are correct! – Usermath Jan 25 '18 at 1:26