“Pseudo-Cauchy” sequences: are they also Cauchy?

I tried to prove something but I could not, I don’t know if it’s true or not, but I did not found a counterexample.

Let $(a_n)$ be a sequence in a general metric space such that for any fixed $k \in \mathbb N$, we have $|a_{kn} – a_n| \to 0$. Is $a_n$ necessarily a Cauchy sequence?

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The additional question about adding the condition that the sequence is bounded caught my attention. (It is in Daniel’s comment under Brian’s answer.)

First I was able to get an example only for $k=2$:
$a_{2^x+y}=\frac y{2^x}$ for $0\le y<2^x$.

When trying to find an example working for all $k$’s I got stuck for some time, so I tried to modify Brian’s example:
$a_n=\sin\left(\frac\pi2 \lg\lg n\right)$.

To show that it fulfills the requirements, we can use $|\sin x-\sin y|\le |x-y|$ and the same reasoning as Brian did. (In detail: $|a_{kn}-a_n| = |\sin(\frac\pi2\lg\lg kn)-\sin(\frac\pi2\lg\lg n)| \le \frac\pi2(\lg\lg kn – \lg\lg n)$, an the RHS converges to 0, as shown in Brian’s answer. Thus $|a_{kn}-a_n|\to 0$.)

But this sequence has a subsequence convergent to 0 (for $n=2^{2^{2k}}$) and a subsequence convergent to 1 (for $n=2^{2^{4k+1}}$). Hence it is not convergent and, consequently, not Cauchy.

I hope I did not miss some mistake there.

It need not be a Cauchy sequence; here’s a counterexample in $\mathbb{R}$.

Let $a_n = \ln\ln n$. Then $|a_{kn} – a_n| = \ln\ln kn – \ln\ln n = \ln\frac{\ln kn}{\ln n} = \ln\frac{\ln k + \ln n}{\ln n} = \ln\left(1 + \frac{\ln k}{\ln n}\right) \to 0$ as $n \to \infty$, but the sequence $\langle a_n \rangle_n$ is unbounded and hence not Cauchy.