# An example of a divergent sequence $\{x_n\}$ but with $\lim_{n\to\infty} |x_{n+p} - x_n| = 0$ for any $p \in \Bbb N$

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.

## 4 Answers

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$$\forall \epsilon>0\exists N\forall p>0\forall n>N(|x_{n+p}-x_n|<\epsilon).$$

• Oh, i see. That subtle point makes a lot of difference. Thank you for clarification – roman Dec 26 '18 at 14:10

The harmonic series works. We have that $$|x_{n+p} - x_n| = \sum_{k=n+1}^{n+p} \frac{1}{k} \leq \frac{p}{n+1} \xrightarrow[n\to\infty]{} 0$$

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 $$p$$.

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