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

calculusconvergence-divergenceexamples-counterexampleslimitssequences-and-series

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.

Best Answer

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

By the way, "Moreover intuitively $|x_{n+p}−x_n|=0$ implies that the sequences eventually turns into a constant which must be convergent" is false; consider for example $x_n=(-1)^n$, $p=2$.