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I should probably not be asking any more questions because it's too late, but this one fascinated me.

If $\ (a_n)_{n\in\mathbb{N}}$ is a strictly increasing sequence of positive real numbers such that

$$\displaystyle\lim_{n\to\infty} a_n = \infty\quad \text{and}\quad \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1,$$

then is the set:

$$ X = \left\{ \frac{a_i}{a_j}: i,j\in\mathbb{N} \right\} $$

dense in $\ [0,1]\ $ (and therefore also dense in $\ \mathbb{R}^+\ ) ?$

$$$$ Without the requirement $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1,\ $ the answer would be "no". For example, consider the sequence $\ a_n=2^n.$

So it is this requirement that makes this question interesting.

Adam Rubinson
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  • Mentioning where you found it, and also mentioning the counterexample without the requirement $\lim_{n \to \infty}\frac{a_{n+1}}{a_n} = 1$ will be helpful for context, thank you. – Sarvesh Ravichandran Iyer May 22 '21 at 13:43
  • I didn't find it: I came up with the question myself. And I've edited the question to include a ounterexample, as requested. Also, the person who downvoted because the question seems irrelevant to the community: I don't see how my question is any less valuable than the next real analysis question. Yes it not immediately applicable to real life, but that's the nature of real analysis: we are interested in the methods of solutions to nontrivial problems because it helps improve our problem-solving skills, which is what maths is all about. – Adam Rubinson May 22 '21 at 13:56
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    But you must have seen some kind of phenomena before you came up with this question , right? I mean, maybe a particular instance of it , or a less general result? All I'm asking for is some context, which is : motivation/source/attempt/your level or background. If you add these things it makes the question more interesting as well as attractive to others. To give you an example : say I take the sequence $a_n = \ln n$, can you frame an argument that works for this particular sequence? – Sarvesh Ravichandran Iyer May 22 '21 at 14:07
  • I'm not sure what you mean. Sometimes I just sit there and come up with interesting questions. I'm not at university or anything (although I was 10 years ago) - I guess I have a general interest in maths, in particular real analysis and always have done, so this is more recreational for me. As for motivation behind the question, I first asked myself the question with the requirement $\ \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\beta>1,\ $ and the answer seemed fairly clear "no" to me. Then I asked myself, what about prime numbers, which I know have the property – Adam Rubinson May 22 '21 at 14:45
  • $\ \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1,\ $ the answer isn't clear to me. And then the question naturally followed from there. As for $\ a_n= \ln n,\ $ that shouldn't be too difficult to prove because we can use the property $\ \ln (kn) = \ln(k) + \ln(n).\ $ But for example, something like $\ a_n=n^3\ $ is less obvious to me whether or not it will work and I want a general method to prove it for all functions with the property $\ \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1. $ – Adam Rubinson May 22 '21 at 14:47
  • Now that I think of it, prime numbers should be just as hard to prove as $\ \ln $ because $\pi(n)\approx\ \frac{n}{\ln(n)},\ $ which has similar global properties to $\ \ln n$. Anyway, a general proof (like the one in the only current answer) would answer all these questions at once... But it sounds like you don't believe me and you think this is my homework or I found it from a textbook, to which there's nothing I can say that can prove that you're right or wrong... So you're welcome to believe what you want. – Adam Rubinson May 22 '21 at 14:56
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    It is not that I don't believe you, of course I do. But I want to say that this question , for it to stay, must have context. You have read "How to ask a good question", and I thank you very much for mentioning the source (self-thought) and counterexample. Now, what I realized following some thought, is that $X$ is closed under multiplication and contains elements arbitrarily close to $1$. The reason I asked you to take $\ln$, is because I thought that you might benefit from looking at the additive version of this problem, which is : set closed under addition, containing ... – Sarvesh Ravichandran Iyer May 22 '21 at 15:32
  • ... elements arbitrarily close to zero. Can you think of such sets? Perhaps you can try to think if these are dense in the reals? Now, when I think about that particular condition, what strikes me is the Riemann Rearrangement theorem and its proof. If you think along these lines and have an attempt, you can add it to the post. It's not that I don't want to help, I try to keep questions afloat. But there is no doubt that this question can be enhanced, and that's all I'm asking for. I've even given you a direction to work in, which may be beneficial. – Sarvesh Ravichandran Iyer May 22 '21 at 15:33
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    I know what the Riemann Series theorem is, but I do not know it's proof. But I will bear that in mind when I revisit this in the future. – Adam Rubinson May 22 '21 at 16:22
  • There is a similar question (could be the same question) here as well. – Sarvesh Ravichandran Iyer May 23 '21 at 08:48
  • In fact, I talked about multiplicatively closed sets : here is another kind of density result relating to such sets. The basic idea is always the same : being closed under multiplication and having elements arbitrarily close to $1$. – Sarvesh Ravichandran Iyer May 23 '21 at 09:02

1 Answers1

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The answer is yes. Here is a sketch:

Let $0< a <b < 1 $ be arbitrary.

Since $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ there exists some $N$ such that for all $n>N$ we have $$ \frac{a_{n+1}}{a_n} < \frac{b}{a} \,. $$

Since $a_n$ is increasing we have

$$ \frac{a}{b}< \frac{a_{n}}{a_{n+1}} < 1 \forall n >N \qquad (*) $$

Next, using again $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{a_{n+1}}=1$ there exists some $M>N$ such that $$ \frac{a_{M}}{a_{M+1}}> b $$

Finally, since $$ \lim_k \frac{a_{M}}{a_{M+k}} =0 $$ there exists some $k \geq 2$ such that $$ \frac{a_{M}}{a_{M+k}} < b \,. $$ Pick the minimal such $k$. Then $$ \frac{a_{M}}{a_{M+k-1}} \geq b \\ \frac{a_{M+k-1}}{a_{M+k}} > \frac{b}{a} \qquad \mbox{ by } (*) \,. $$

Multiplying gives $\frac{a_{M}}{a_{M+k}} >a$ which completes the proof.

P.S. Intuitivelly the solution is as follows: for $n$ large enough $\frac{a_n}{a_{n+1}}$ is arbitrary closed to $1$, but smaller than one. Multiplying consecutive terms of this form, you get telescopic products converging to $0$ is very small steps.

N. S.
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