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.
 A: 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 $x_n=(-1)^n$, $p=2$.
A: The harmonic series works. We have that $$|x_{n+p} - x_n| = \sum_{k=n+1}^{n+p}  \frac{1}{k} \leq \frac{p}{n+1} \xrightarrow[n\to\infty]{} 0$$
A: 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 $p$.
A: Two more examples: $x_n=\log n$ (more or less the same as in the previous answers), $x_n=\sqrt{n}$.
