Let $(x_n)$ represent an arbitraty sequence that satisfies $$\lim_\limits{n\to \infty}|x_{n+1}-x_n|=0$$ (i) Is it possible to say that $(x_n)$ is a Cauchy sequence?

(ii) Instead, suppose now that $(x_n)$ satisfies $\forall p\in \mathbb{N},$ $$\lim_\limits{n\to \infty}|x_{n+p}-x_n|=0$$ An then answer the same question.

$$|x_{n+1}-x_n|\ge |x_{n+1}|-|x_n|$$So, by taking limits on both sides$$\lim_\limits{n\to \infty}|x_{n+1}-x_n|\ge \lim|x_{n+1}|-\lim|x_n|\implies \lim|x_{n+1}|\le\lim|x_n|$$ Here i see that if $\lim|x_{n+1}|=\lim|x_n|$, then $(|x_n|)$ converges. If $\lim|x_{n+1}|<\lim|x_n|$, then $(|x_n|)$ diverges. Well, if it is the first situation, than i cant say anything about $(x_n)$. If it is the other one, i suppose then $(|x_n|)$ divergent $\implies$ $(x_n)$ divergent. So i am in a situation where i dont know how to properly infer anything about $(x_n)$. After some time i came to think about $x_n=\sqrt{n}$, and it seemed to be a plausible counter-example that would then proove (i) and (ii) to be false since $\lim|\sqrt{n+1}-\sqrt{n}|=0$ and it is not Cauchy. Still i feel like i'm missing something. Can anyone give me a hand on that exercise? Grateful for any help!

  • $\begingroup$ In the second case, you have an inequality between limits, and the conclusion is there's no limit? $\endgroup$
    – Bernard
    Jul 28, 2017 at 16:35
  • 2
    $\begingroup$ You're right. $x_n=\sqrt{n}$ satisfies $(i)$ and $(ii).$ However it's not a Cauchy sequence. $\endgroup$
    – mfl
    Jul 28, 2017 at 16:43

1 Answer 1


(ii) does not imply Cauchy.

Let $$ x_n=\log(n+1) $$ Then for each $p$, $$ \begin{align} \lim_{n\to\infty}|x_{n+p}-x_n| &=\lim_{n\to\infty}\log\left(\frac{n+p+1}{n+1}\right)\\ &=0 \end{align} $$ However, $x_n$ is not Cauchy.

Since (ii) does not imply Cauchy, (1) does not imply Cauchy since it is (ii) with $p=1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .