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Problem: Show that $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$ can never be an integer $\forall \ n \ge 1$.

Now, I know that there are quite a few proofs available. In particular, the $2$ solutions that I am familiar with uses Bertrand's postulate (which in my opinion is a considerably advanced tool), while the other solution involves a neat trick which allows us to derive a contradiction. However, I would like to check the validity of a proof, also by contradiction, which might seem a little too simple; I am worried that I have made some oversights somewhere. The details of the "proof" are as follows:

Let $T \subset \mathbb{N^+} \setminus \{1\}$ be a subset of $\mathbb{N^+}$ such that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{t} \ \in \mathbb{Z^+} \ \forall \ t \in T,$ $T$ is non-empty. Then, by the Well-Ordering Principle, $T$ has a least element, $t_0 \in T$.

Thus, $$1+\frac{1}{2}+...+\frac{1}{t_0} \in \mathbb{Z^+} \iff \frac{t_0!+\frac{t_0!}{2}+...+\frac{t_0!}{t_0}}{t_0!} \in \mathbb{Z^+}.$$

Now, there exists $k \in \mathbb{Z^+}$ such that $2^k\mid\mid t_0!$.

Hence, we must have $2^k \mid t_0!+\frac{t_0!}{2}+...+\frac{t_0!}{t_0} \Rightarrow 2^k \mid \frac{t_0!}{2}+...+\frac{t_0!}{t_0} $. Next, let $m$ be the largest positive integer such that $2m \le t_0$. Clearly, $m < t_0$.

Also, we must have: $2^k \mid \frac{t_0!}{2}+\frac{t_0!}{4}+...+\frac{t_0!}{2m}$. ( This is because $2^k$ trivially divides $\frac{t_0!}{l}$ if $l$ is odd. ) Hence, we obtain:

$$2^k \mid \frac{t_0!}{2}\left(1+\frac{1}{2}+...+\frac{1}{m}\right)$$

But, clearly, $2^{k-1} \mid\mid \frac{t_0!}{2}$, hence we must have that $2 \mid 1+\frac{1}{2}+...+\frac{1}{m} \Rightarrow 1+\frac{1}{2}+...+\frac{1}{m} \in \mathbb{Z^+}$, which contradicts the minimality of $t_0$ in $T$!

Would appreciate if anyone could help check my "proof" and point out any fallacious ideas if there are any!

Hello
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    I'm not sure about your last step. When you conclude that 2 | 1+1/2+...+1/m, you seem to be assuming that 1+1/2+...+1/m is an integer, which is what you're trying to prove? – AlephNull May 22 '20 at 14:51
  • If $2^k|t_0! + etc \implies 2^k|etc.$ leads to a contradiction. Fine. so $2^k\not \mid t_0! + etc.$ How is that a contradiction? $False \implies False$ is not a contradiction. So you have $2^k|t_0!$ bue $2^k\not\mid t_0! + etc.$. That not a contradiction. – fleablood May 22 '20 at 15:11
  • What's the difference between double bar, $a||b$, and the single bar, $|$, which means $a$ divides $b$? What does $a||b$ mean? – fleablood May 22 '20 at 15:17
  • @fleablood In the context of this question it refers to the largest power of $k$ such that $2^k \mid t_0!$. – Hello May 22 '20 at 15:18
  • But anyways, the others have already pointed out a glaring flaw in my proof. – Hello May 22 '20 at 15:19

1 Answers1

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You have the statement

$$2^k \mid \frac{t_0!}{2}\left(1+\frac{1}{2}+...+\frac{1}{m}\right)$$ which means that the RHS is a positive integer and a multiple of $2^k$.

However, of the two factors on the $\text{RHS}$, you know the factor on the left is an integer, but you can't automatically infer that the factor on the right is also an integer.

quasi
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  • But since $2^{k-1} \mid\mid \frac{t_0!}{2}$, in order for the RHS to be a multiple of $2^k$, isn't it necessary for us to have $2 \mid 1+\frac{1}{2}+...+\frac{1}{m}$? Otherwise the R.H.S. is only a multiple of $2^{k-1}$ but not $2^k$? – Hello May 22 '20 at 15:00
  • Sort of. The extra factor of $2$ needs to divide the numerator of the factor on the right, but that doesn't automatically imply that the factor on the right is an integer. When reduced to lowest terms, the factor on the right potentially could have an odd denominator greater than $1$. – quasi May 22 '20 at 15:01
  • Hmmm, so $2 \mid 1+\frac{1}{2}+...+\frac{1}{m} \nRightarrow 1+\frac{1}{2}+...+\frac{1}{n} \in \mathbb{Z^+}$? – Hello May 22 '20 at 15:03
  • Sorry, how does this work out? Doesn't the notion of divisibility only apply to integers? – Hello May 22 '20 at 15:04
  • Yes, the RHS is an integer, but that doesn't mean both factors of the RHS are integers. – quasi May 22 '20 at 15:05
  • For example $4{,\mid,}\left(30{,\cdot,}\frac{2}{5}\right)$, but $\frac{2}{5}$ is not an integer. – quasi May 22 '20 at 15:08
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    We have $2^{k-1} \ || \ t_0!/2$, and $2^k \ | \ (t_0!/2) x$ where $x:=1+1/2+\ldots+1/m$ may or may not be an integer. What if $x=2^{k+1}/(t_0!)$ for example? Then the conditions are satisfied, but $x$ is not an integer. – AlephNull May 22 '20 at 15:08
  • Of course, my bad! This is indeed my oversight! Thanks! – Hello May 22 '20 at 15:09