What is an example of a sequence in $\mathbb R$ with this property that is not Cauchy?

I know that Cauchy condition means that for each $\varepsilon>0$ there exists $N$ such that $d(x_p,x_q)<\varepsilon$ whenever $p,q>N$.

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    $\begingroup$ Hint: look for a sequence $x_n$ with $x_n \rightarrow \infty$. $\endgroup$ – Pete L. Clark Feb 9 '12 at 4:25
  • $\begingroup$ Of course that condition becomes an equivalence condition for the Cauchy criterion in some spaces such as $\mathbb{Q}_{p}$, but not in $\mathbb{R}$ where cumulative effect is non-negligible as you can see several examples from the answers below. $\endgroup$ – Sangchul Lee Feb 9 '12 at 5:06
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    $\begingroup$ The only condition that this imposes on the sequence is that its limit set is an interval. $\endgroup$ – Did Feb 9 '12 at 6:11
  • $\begingroup$ It is true if $d$ is ultrametric (e.g. in $p$-adic analysis). $\endgroup$ – Watson Nov 28 '18 at 13:00

Let $$ s_{n}:= \sum_{i=1}^{n} \frac{1}{i}. $$ Then $$ s_{n+1}-s_{n} = \frac{1}{n+1} \to 0 $$ as $n\to \infty$. However, this sequence is not Cauchy, as for any $\varepsilon > 0$, there is no $N$ such that $ |s_{m}-s_{n}| < \varepsilon $ for all $n,m\ge N$. In fact, $$ \lim_{m\to\infty} |s_{m}-s_{n}| = \infty, $$ for any $n$.

The "for all" quantifier on $m$ and $n$ is what confuses people about Cauchy sequences.


$$\textstyle 0,\,{1\over 2},\, 1,\, {2\over 3},\,{1\over 3},\,{0},\,{1\over 4},\,{2\over 4},\,{3\over 4},\,1,\,{4\over 5},\,{3\over 5},\,{2\over 5},\,{1\over 5}, \,0, \ldots$$

  • $\begingroup$ Nice example. Wish I had thought of it, mine is boring. $\endgroup$ – André Nicolas Feb 9 '12 at 5:59
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    $\begingroup$ One may even modify this to get the whole real line as limit set. $\endgroup$ – Did Feb 9 '12 at 6:09
  • $\begingroup$ Brilliant observation @DidierPiau $\endgroup$ – Inquest Feb 9 '12 at 12:14
  • $\begingroup$ I think $x_n=\sqrt n\sin\log n$ is also an example dense in the real line. $\endgroup$ – Jonas Meyer Jun 4 '13 at 19:00

The standard example is the sequence $(s_n)$ of partial sums of the harmonic series. Formally, $$s_n=\sum_{k=1}^n \frac{1}{k}.$$

Note that $d(s_n,s_{n+1})=\frac{1}{n+1}$. It is clear that $d(s_n, s_{n+1})\to 0$ as $n\to\infty$.

But the sequence $(s_n)$ is not Cauchy. For given any $m$, we can find $n$ such that $d(s_m,s_n)$ is arbitrarily large. This is because the sequence $(s_n)$ diverges to infinity. We omit the proof, since you have likely already seen a proof that $\sum_{k=1}^\infty \frac{1}{k}$ diverges.

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    $\begingroup$ +1 because this is exactly what I was going to say, until I noticed that someone beat me to it. This question is pretty much the same thing as asking: If the terms of a series approach $0$, why doesn't that imply that it converges. $\endgroup$ – Michael Hardy Feb 9 '12 at 4:31
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    $\begingroup$ @Michael: the given question is precisely equivalent to your alternative version, in fact. $\endgroup$ – Pete L. Clark Feb 9 '12 at 4:36

Another example in the line with the usual metric is $x_n = \sqrt{n}$.


Let the metric space be $\mathbb R$ with respect to the usual Euclidean distance metric $|.|$ , consider the

sequence $x_n:=\log n , \forall n \in \mathbb Z^+ $ , then $d(x_{n+1}, x_n)=|\log(n+1)-\log n|=|\log\big(1+\dfrac1n\big)| \to \log 1=0$ as $n \to \infty$ , but off-course

$(\log n)_{n=1}^\infty$ is not a bounded sequence , hence is not Cauchy


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