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Let

$$S_m(n) = \frac{1}{m} + \frac{1}{m+1} + \dots + \frac{1}{m+n}$$

where $m,n \in \mathbb{N}$.

Now its a fact that $S$ is never an interger $\forall \ m,n \in \mathbb{N}$. And in fact, the denominator of $S$ is always even in its simplest form, as proven here.

Now, let $S_m(n) = \frac{a}{2^x \cdot b}$ where $a, b$ are two co-prime odd natural numbers.

Define $p^k \mid \mid q$ such that $p^k \mid q$ but $p^{k+1} \nmid q$.

If $2^y \mid \mid (m+n+1)$, prove that $x \ne y \ \forall \ m,n \in \mathbb{N}$.

Or provide a contradictory example.

Nick Larry
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  • There is actually a name for the $||$ symbol you introduce, it is called the 2-adic absolute value. Usually written as $|q|_p=k$ instead of $p^k||q$. – student91 Sep 12 '22 at 14:17
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    Oh, I wasn't aware. I'm still an elementary student. Thanks for informing. But ig I'll just let it be for now. – Nick Larry Sep 12 '22 at 14:25
  • Maybe you can prove the not-so-obvious fact that $x = \max{ k : p^k|m+i$ for some $0\leq i\leq n}$ and use that $y$ cannot be the same as this maximum because $2$ is a special number? – student91 Sep 12 '22 at 14:46
  • I think that this fact is the key here. Because when you try to pick $m \ & \ n$ such that $x = 2$, say, then if you use only 2 terms, the next even number gives $y = 1$ or if you take 4 terms instead, $y = 3$. No matter what $x$ you choose, you'll end up getting one more or one less factor of $2$ in the next term at best. – Nick Larry Sep 12 '22 at 14:57
  • @student91I think I got it! I'll write up the proof soon. – Nick Larry Sep 12 '22 at 15:28
  • Nice! Note however, that this fact is stronger than the original question you are trying to answer. I guess that is nice. – student91 Sep 12 '22 at 15:38
  • Yeah, proving this fact alone solves the oroginal problem. But its still nice to finish what I started. – Nick Larry Sep 12 '22 at 16:20
  • @student91I've written up the proof. I'd like your feedback. Thanks. – Nick Larry Sep 12 '22 at 17:18

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Okay, so the first thing we need to prove is that, $$x = \max\{k:2^k \mid \mid m + i, \ 0 \le i \le n\}$$

Let $H = \{m, m+1, m+2, \dots, m+n\}$.

So let's write $S_m(n)$ as a fraction. For that we need to define: $$t_r = \frac{(m+n)!}{(m+r) \cdot m!} = m \cdot (m+1) \ldots (m+r-1) \cdot (m+r+1) \ldots (m+n)$$ Basically, $t_r$ is the product of all the terms from $H$ except for the $(m+r)$ term.

Now, we can write $$S_m(n) = \frac{t_0 + t_1 + t_2 + \dots + t_n}{m \cdot (m+1) \dots (m+n)}$$

Now let $x = \max\{k:2^k \mid \mid m + i, \ 0 \le i \le n\}$, say the $i$ which gives the max $x$ is $i = p$.

This can easily be proven that there is a unique maximum value of $x$ and it occurs only once in the set above. Because the set of any two consecutive multiples of $2^x$ contains a multiple of $2^{x+1}$, thereby contradicting the fact that $x$ is the max value of the set.

Coming back to $S_m(n)$, note that in the numerator, all the terms except for $t_p$ contain a factor of $2^x$. And $t_p$ contains all the multiples of $2$ from the denominator except for $2^x$.

Now consider a random $t_r, r \ne p$ such that $(m+r)$ has a factor of $2^w, w < x$. Since all other terms in the numerator contain $(m+r)$ and $t_r$ contains a factor of $2^x, x>w$ $\implies 2^w \mid \sum{t_j}$.

Since its true for all multiples of $2$ in the denominator (except for $2^x$), they can all be cancelled out. Then we'd be left with a factor of $2$ in all the terms except $t_p$, as all factors of $2$ are cancelled out from $t_p$ but $x > w$ for all the other terms.

Therefore, in $S_m(n) = \frac{a}{2^x \cdot c}$ (its simplest form). We have $x = \max\{k:2^k \mid \mid m + i, \ 0 \le i \le n\}$.

Now we can easily say, $y$ such that $2^y \mid \mid (m+n+1)$ is not equal to $x$.

If $x = y$, $(m+p)$ and $(m+n+1)$ are two consecutive multiples of $2^x$ and therefore one of them is an even multiple of $2^x$ and so either $x = x+1$ or $y = x+1$ $\implies x \ne y$.

Note: I know its a really wordy (and probably stupid) proof. And also that its a major overkill for the original question which I was trying to solve. But at least I'd be able to sleep better now. Thanks for reading.

Nick Larry
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  • I believe you do not yet have the part where $2^{x+1}$ does not divide $\sum t_j$. All else looks great! Hint: $\frac{1}{m(m+1)\cdots(m+n)}\sum_{j\neq r} t_j$ is (a sum of "even" numbers and therefore) "even" and $\frac{t_r}{m(m+1)\cdots(m+n)}$ is "odd"; odd+even=odd. To make "even" and "odd" precise, work in $\mathbb{Q}_2={a/b : 2\nmid b}$ instead of $\mathbb{Z}$. But I'm not very confident in that area tbh. – student91 Sep 12 '22 at 19:20
  • Oh wait you did say something in that regard – student91 Sep 12 '22 at 19:23