I tried to prove something but I could not, I don't know if it's true or not, but I did not found a counterexample.

Let $(a_n)$ be a sequence in a general metric space such that for any fixed $k \in \mathbb N$, we have $|a_{kn} - a_n| \to 0$. Is $a_n$ necessarily a Cauchy sequence?


The additional question about adding the condition that the sequence is bounded caught my attention. (It is in Daniel's comment under Brian's answer.)

First I was able to get an example only for $k=2$:
$a_{2^x+y}=\frac y{2^x}$ for $0\le y<2^x$.

When trying to find an example working for all $k$'s I got stuck for some time, so I tried to modify Brian's example:
$a_n=\sin\left(\frac\pi2 \lg\lg n\right)$.

To show that it fulfills the requirements, we can use $|\sin x-\sin y|\le |x-y|$ and the same reasoning as Brian did. (In detail: $|a_{kn}-a_n| = |\sin(\frac\pi2\lg\lg kn)-\sin(\frac\pi2\lg\lg n)| \le \frac\pi2(\lg\lg kn - \lg\lg n)$, an the RHS converges to 0, as shown in Brian's answer. Thus $|a_{kn}-a_n|\to 0$.)

But this sequence has a subsequence convergent to 0 (for $n=2^{2^{2k}}$) and a subsequence convergent to 1 (for $n=2^{2^{4k+1}}$). Hence it is not convergent and, consequently, not Cauchy.

I hope I did not miss some mistake there.

  • $\begingroup$ Very nice! Much cleverer than the blind alleys I kept running down. $\endgroup$ – Brian M. Scott Aug 9 '11 at 6:24
  • $\begingroup$ Martin in this case your inequality it´s in the reverse order, so property it´s not meet in this case $\endgroup$ – Daniel Aug 9 '11 at 16:30
  • $\begingroup$ @Daniel: I'm not sure what you mean. I get from the above $|a_{kn}-a_n|\le \frac\pi2 ( \lg \lg kn - \lg \lg n)$, and the RHS tends to 0. $\endgroup$ – Martin Sleziak Aug 9 '11 at 16:39
  • $\begingroup$ Oh sorry , I was wrong, great example, :D!!!! $\endgroup$ – Daniel Aug 16 '11 at 2:12

It need not be a Cauchy sequence; here’s a counterexample in $\mathbb{R}$.

Let $a_n = \ln\ln n$. Then $|a_{kn} - a_n| = \ln\ln kn - \ln\ln n = \ln\frac{\ln kn}{\ln n} = \ln\frac{\ln k + \ln n}{\ln n} = \ln\left(1 + \frac{\ln k}{\ln n}\right) \to 0$ as $n \to \infty$, but the sequence $\langle a_n \rangle_n$ is unbounded and hence not Cauchy.

  • $\begingroup$ @jspecter: Are you sure? That was my first thought, but that’s roughly $a_n=\ln n$, for which $|a_{kn}-a_n|=\ln kn - \ln n = \ln k$ does not go to $0$. That’s why I ended up making it grow even more slowly. $\endgroup$ – Brian M. Scott Aug 5 '11 at 4:59
  • $\begingroup$ good counterexample $\endgroup$ – Daniel Aug 5 '11 at 14:15
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    $\begingroup$ And whatit happens if the sequence is bounded? (at least in the real numbers) $\endgroup$ – Daniel Aug 5 '11 at 18:26
  • $\begingroup$ @Daniel: You accidentally created a duplicate account; I've merged it with your main account. Also, the post you just made a few minutes ago was as an answer to this question, when it really ought to have been a comment (you were unable to post a comment because your duplicate account didn't have enough reputation points to comment. I've now converted that post to a comment on Brian's answer here; I assume this is what you intended. $\endgroup$ – Zev Chonoles Aug 5 '11 at 18:39
  • $\begingroup$ Isn't it nice? I upvoted Martin's answer, thus giving you both badges: him the Nice Answer badge, and you the Populist gold badge. $\endgroup$ – Asaf Karagila Feb 25 '12 at 10:48

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