If $a_{n+f(n)}-a_n\rightarrow 0$ as $n\rightarrow\infty$ for every function $f$ from the positive integers to the positive integers, does it follow that $(a_n)$ is convergent?
I think the answer is 'yes' but I'm having a hard time proving it. I have tried to show that it is Cauchy but to no avail.
Any help is appreciated, thanks
Assume for the sake of contradiction that $(a_n)$ is not Cauchy. Then there is some $\varepsilon > 0$ such that for all $n \in \mathbb{N}$ there are $k,m > n$ satisfying $|a_k-a_m| \ge \varepsilon$.
For each $n = 1,2,\ldots$ choose some $k(n),m(n)$ as above and denote two functions $f,g \colon \mathbb{N} \to \mathbb{N}$ by $f(n) = k(n)-n$ and $g(n) = m(n)-n$. Then $|a_{n+f(n)}-a_{n+g(n)}| \ge \varepsilon$, but on the other hand $$ |a_{n+f(n)}-a_{n+g(n)}| \le |a_{n+f(n)}-a_{n}| + |a_{n+g(n)}-a_{n}| \xrightarrow{n \to \infty} 0, $$ which yields a contradiction.