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Find an example of a divergent sequence $\{x_n\}$ such that $\forall p \in \Bbb N$: $$ \lim_{n\to\infty} |x_{n+p} - x_n| = 0 $$

The way the problem is stated suggests that the sequence exists, however I wasn't able to find such a sequence. Moreover it seems impossible since from the definition of a limit above: $$ \lim_{n\to\infty} |x_{n+p} - x_n| = 0 \stackrel{\text{def}}{\iff} \forall \epsilon>0\ \exists N\in\Bbb N: \forall n > N \implies |x_{n+p} - x_n| < \epsilon $$

Which is nothing but the fact that $x_n$ is fundamental and therefore convergent by the Cauchy Criterion.

Moreover intuitively $|x_{n+p} - x_n| = 0$ implies that the sequences eventually turns into a constant which must be convergent.

Could someone please provide an example of such a sequence in case it indeed exists? There may be some esoteric sequences I'm not aware of.

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An example that's actually equivalent to the other examples given: $x_n=\log(n)$.

The reason the given condition does not imply that $(x_n)$ is a Cauchy sequence is that $p$ is fixed.. That is, the order of the quanitifers is different; the given condition is $$\forall p>0\forall \epsilon>0\exists N\forall n>N(|x_{n+p}-x_n|<\epsilon),$$while saying $(x_n)$ is Cauchy is the stronger condition$$\forall \epsilon>0\exists N\forall p>0\forall n>N(|x_{n+p}-x_n|<\epsilon).$$

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  • $\begingroup$ Oh, i see. That subtle point makes a lot of difference. Thank you for clarification $\endgroup$ – roman Dec 26 '18 at 14:10
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The harmonic series works. We have that $$|x_{n+p} - x_n| = \sum_{k=n+1}^{n+p} \frac{1}{k} \leq \frac{p}{n+1} \xrightarrow[n\to\infty]{} 0$$

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Let $x_n = \sum_{i = 1}^n \frac{1}{i}$. Then $|x_{n+p} - x_n| = \sum_{i = n+1}^{n+p} \frac{1}{i} \leq \frac{p}{n} \rightarrow 0$ as $n \rightarrow \infty$ for any $p$.

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Two more examples: $x_n=\log n$ (more or less the same as in the previous answers), $x_n=\sqrt{n}$.

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