 Solution

Yes. The arbitrary variable-lag condition forces the Cauchy sequence property. Suppose otherwise. Then some ${\epsilon }_0>0$ has the property that, for every $N$, there exist $p,q\ge N$ with $|a_p-a_q|\ge {\epsilon }_0$. In particular, for each $n$, choose such $p,q\ge n$. By the triangle inequality, at least one satisfies $|a_p-a_n|\ge {\epsilon }_0/2$ or $|a_q-a_n|\ge {\epsilon }_0/2$. That index must be strictly greater than $n$, since the difference at $n$ itself is zero.

It follows that the following positive integer is defined for every $n$:For this particular function, $|a_{n+f(n)}-a_n|\ge {\epsilon }_0/2$ for every $n$, contradicting the assumed limit zero. Therefore $(a_n)$ is a Cauchy sequence, and the completeness of the real numbers proves it converges to a finite real limit. This is convergence from arbitrary variable-lag increments; the quantifier over every function is much stronger than the fixed-lag condition in the previous part.