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Let $\{a_n\}$ be a decreasing sequence and $a_n>0$ for all $n$. If $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ and $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$, how to prove or disprove that $\displaystyle\lim_{n\to\infty}\frac{a_{3n}}{a_n}=\frac{1}{3}$ ?

Thank you.

Bless
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  • 6
    What did you try and what is your progress so far? – Sungjin Kim Mar 20 '17 at 16:56
  • Is it true that powers of 2 and powers of 3 come arbitrarily close to each other? – Aritro Pathak Mar 20 '17 at 18:13
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    You can't disprove this since $a_n = n^{-1}$ exhibits an explicit example of $\lim_{n \to \infty} a_{3n}/a_n = 1/3$, so the only options remaining is to 1。 prove it, or 2. show that $\lim_{n \to \infty} a_{3n}/a_n = 1/3$ may or may not hold by exhibiting another example. – Henricus V. Mar 20 '17 at 20:06
  • I think if in the sequence $a_n$, we had $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, it is not possible that $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$ because: – Amin235 Mar 20 '17 at 20:44
  • \begin{eqnarray} \displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}&=& \displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\times \frac{a_{n+2}}{a_{n+1}}\times \cdots \times \frac{a_{2n}}{a_{2n-1}} \ && \ &=&\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\times \displaystyle\lim_{n\to\infty}\frac{a_{n+2}}{a_{n+1}}\times \cdots \times \displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_{2n-1}}\ &&\ &=&1\times 1\times \cdots \times 1\ &&\ &=&1 \end{eqnarray} – Amin235 Mar 20 '17 at 20:44
  • @Amin235 What about $a_n=1/n$? (Your rule for taking limits holds for a finite number of factors. In your case the number of factors grow with $n$) – mickep Mar 20 '17 at 20:58
  • @mickep Is there a proof that my rule for taking limits does not hold for infinity number of factors? – Amin235 Mar 20 '17 at 21:05
  • @Amin235 Yes, the counterexample just given serves as a proof. – mickep Mar 20 '17 at 21:09
  • @mickep you right, but if for the sequence $a_n=\frac{1}{n}$, we compute the following equation $$ \displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\times \frac{a_{n+2}}{a_{n+1}}\times \cdots \times \frac{a_{2n}}{a_{2n-1}} $$ why the answer $1$ for the above limit is incorrect. I mean, what error there is in my discussion. This question is so important for me because I have used of this argument for obtaining the limit values of recursive sequences. Thanks for any help. – Amin235 Mar 20 '17 at 21:22
  • There's no operation you can do to limits "by default": any property of limits that we use beyond the definition, we use because we've proven that it works.

    We can prove that $\lim a_n b_n = (\lim a_n)(\lim b_n)$, and we can use this inductively to break up the limit of a $k$-fold product of limits, for any fixed $k$. This argument doesn't let us break up an $n$-fold product, where the $n$ is the same as the variable going to $\infty$.

     – Misha Lavrov Mar 20 '17 at 21:52
  • @Misha For example please see my answer of this post. now by the taking limit on the $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}$, we can find some formula based on the parameters $\alpha_1$ and $\alpha_2$ and by numerical simulation we can see that is correct. – Amin235 Mar 20 '17 at 22:10

2 Answers2

14

The limit of $\frac{a_{3n}}{a_n}$ need not be $\frac13$, because it need not exist.

Each $n>0$ can be written uniquely as $2^k + r$, where $2^k \le n < 2^{k+1}$. Define $a_n$ (in terms of these $k$ and $r$) piecewise as follows: $$a_n = \begin{cases} \frac{2^{1 - r/2^{k-1}}}{2^k} & \mbox{if }r < 2^{k-1} \\ \frac{1}{2^k} & \mbox{if }r \ge 2^{k-1}. \end{cases}$$

Essentially, for the first half of the range from $2^k$ to $2^{k+1}$, $a_n$ decreases from $\frac{1}{2^{k-1}}$ to $\frac1{2^k}$ geometrically, by factors of $2^{-1/2^{k-1}}$. For the second half of that range, $a_n$ stays at $\frac1{2^k}$.

It is identically true that $\frac{a_{2n}}{a_n} = \frac12$. Moreover, if $n \ge 2^k$, then $2^{-1/2^{k-1}} \le \frac{a_{n+1}}{a_n} \le 1$, so $\frac{a_{n+1}}{a_n} \to 1$.

However, when $n = 2^k$, $\frac{a_{3n}}{a_n} = \frac14$, while $\frac{a_{9n}}{a_{3n}} = \frac1{2^{5/4}}$, so $\frac{a_{3n}}{a_n}$ does not converge.

You might complain that the sequence $a_n$ is not strictly decreasing. If this is a problem, just replace $a_n$ by $a_n' = a_n\left(1 + \frac1n\right)$. We have:

  • $\frac{a'_{n+1}}{a'_n} = \frac{a_{n+1}}{a_n} \cdot \frac{1+\frac1{n+1}}{1+\frac1n} = \frac{a_{n+1}}{a_n} \left(1 - \frac1{(n+1)^2}\right)$, which still converges to 1.
  • $\frac{a'_{2n}}{a'_n} = \frac{a_{2n}}{a_n} \cdot \frac{1 + \frac1{2n}}{1 + \frac1n} = \frac{a_{2n}}{a_n} \left(1 - \frac1{2n+2}\right)$, which still converges to $\frac12$.
  • $\frac{a'_{3n}}{a'_n} = \frac{a_{3n}}{a_n} \cdot \frac{1 + \frac1{3n}}{1 + \frac1n} = \frac{a_{3n}}{a_n} \left(1 - \frac2{3n+3}\right)$, which still does not converge to anything.
  • Since we already had $a_{n+1} \le a_n$, we now have $a'_{n+1} = a_{n+1} \left(1 + \frac1{n+1}\right) \le a_n \left(1 + \frac1{n+1}\right) < a_n \left(1 + \frac1n\right) = a'_n$.

If the limit $\frac{a_{3n}}{a_n}$ does exist, then we may proceed as in Andras's answer to show that it must equal $\frac13$:

If $\frac{a_{3n}}{a_n} \to c$, then $\frac{a_{3^kn}}{a_n} \to c^k$ for any $k$, but we can bound $\frac{a_{3^kn}}{a_n}$ between $\frac{a_{2^\ell n}}{a_n}$ and $\frac{a_{2^{\ell+1} n}}{a_n}$ for $\ell$ such that $2^\ell < 3^k < 2^{\ell+1}$ (i.e., $\ell = \lfloor k \log_2 3\rfloor$). These two ratios converge to $\frac{1}{2^{\ell}}$ and $\frac1{2^{\ell+1}}$, so we get that $2^{-\ell/k} < c < 2^{-(\ell+1)/k}$. Taking $k$ arbitrarily large, $\frac{\ell}{k} \to \log_2 3$, so $c$ must be $2^{-\log_2 3} = \frac13$.

Misha Lavrov
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  • You say "multiply $a_n$ by a sequence that decays to $1$ sufficiently slowly for this not affect convergence of any of the ratios we care about." But you do not need just the convergence of the ratios, you also need the product to be strictly decreasing. I am convinced by all your argument except that part, which is still unclear. – Ewan Delanoy Mar 20 '17 at 20:39
  • Take $a_n' = a_n (1 - \epsilon_n)$, where $\epsilon_n$ is strictly decreasing and $\epsilon_n \to 0$ very very slowly. We already had $a_{n+1} \le a_n$; now we have $a'{n+1} \le a_n' \cdot \frac{1-\epsilon_n}{1-\epsilon{n+1}}$, and $\frac{1-\epsilon_n}{1-\epsilon_{n+1}} < 1$. – Misha Lavrov Mar 20 '17 at 20:49
  • But you've convinced me that it's worth editing the post to make this argument explicit. – Misha Lavrov Mar 20 '17 at 20:51
  • Thanks for your explanation. I've found a variation of your argument that I prefer : start by putting $a_{2^n}=\frac{1}{2^n}$, $a_{32^n}=b_n\frac{1}{2^{n-1}}$, where $(b_n)$ is any sequence taking values in $[\frac{1}{4},\frac{1}{2}]$. Then you can fill the blanks (taking $\frac{a_{2n}}{a_n}$ to be identically $\frac{1}{2}$ as you do), to satsify all the conditions. – Ewan Delanoy Mar 20 '17 at 20:53
  • Yeah, I started out thinking along those lines, but then decided to give an explicit formula - I'm not sure that made things any clearer. (Also, my $\epsilon_n$ argument in my comment above got its signs confused, so I apologize for that, but I'm fairly sure the argument in my answer is correct.) – Misha Lavrov Mar 20 '17 at 21:06
  • "Sqeezing" $2^\ell < 3^k < 2^{\ell+1}$ is exactly the argument I used in my earlier answer. I guess standard rules of academic writing ask for some credit at that point. – Andreas Mar 21 '17 at 07:26
  • As you like; credited. – Misha Lavrov Mar 21 '17 at 11:40
  • @Misha Thanks ! – Andreas Mar 23 '17 at 07:12
7

This is not an answer, but a longer train of thoughts / conjectures that might lead to a full answer.

It might be an idea to use $\displaystyle\lim_{n\to\infty}\frac{a_{2n}}{a_n}=\frac{1}{2}$ and apply it $r$ times. This gives $$ \displaystyle\lim_{n\to\infty}\frac{a_{2^rn}}{a_n}=\frac{1}{2^r} $$ which holds for all $r$. Now take some $m$ and identify $r$ such that $2^r \leq m < 2^{r+1}$. Then one has $$ \frac{1}{2^r} \geq \displaystyle\lim_{n\to\infty}\frac{a_{m n}}{a_n} > \frac{1}{2^{r+1}} $$

In particular, this holds when $m$ is chosen to be any power $3^p$. This gives a countable infinite set of inequalities of the type above, all of which must hold. They do hold if $$ \displaystyle\lim_{n\to\infty}\frac{a_{3 n}}{a_n}=\frac{1}{3} $$ implying $$ \displaystyle\lim_{n\to\infty}\frac{a_{3^p n}}{a_n}=\frac{1}{3^p} $$

The conjecture is that there is no other way that they can hold.

Andreas
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  • Pretty much the same as Vik78 : I downvote, and will reverse my vote if the author inserts the mention "this is not an answer but only a long comment" – Ewan Delanoy Mar 20 '17 at 19:07
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    The author says it might be and calls it a conjecture. What more do you need? – marty cohen Mar 20 '17 at 19:16
  • @martycohen I need a clear statement that this is not an answer. Just putting "It might be" or "conjecture" at some place in the text means nothing. It took me some time to realize that this wasn't an answer, and a reader with a moderate understanding of mathematics might easily mistake it for one. – Ewan Delanoy Mar 20 '17 at 20:28
  • This is not an answer. I put it also in the text. I'm sorry that longer train of thoughts / conjectures are not welcome. – Andreas Mar 21 '17 at 06:38
  • @Andreas Longer train of thoughts/ conjectures are most welcome with me. It's just that to keep a quality standard here on MSE, I think we should distinguish clearly between a comment however elaborate and an answer however quickly sketched. Answerers often make sketchy answers, either because of lack of time or because they want to force the asker to do some work. Thank you for agreeing with me, I just reversed my vote. – Ewan Delanoy Mar 21 '17 at 06:56
  • Yeah, I did as well. – Vik78 Mar 21 '17 at 07:12
  • @EwanDelanoy I'm happy to hear that thoughts/ conjectures are welcome. I agree that this should have been made clear right away. – Andreas Mar 21 '17 at 07:12