$a_{n+f(n)}-a_{n}\rightarrow0$ implies convergence? $(a_{n})$ is a sequence of reals. Say $a_{n+f(n)}-a_{n}$ tends to 0 as n tends to infinity for every function f from the positive integers to the positive integers. Does this imply that $a_{n}$ is convergent?
 A: Yes :
Pick $f(n)$ such that $|a_{n+f(n)} - a_n| \ge \sup_{m \ge n} |a_m -a_n| - \frac 1 n$.
If $a_{n+f(n)} - a_n$ tends to $0$, then the sequence is a Cauchy sequence :
If $\varepsilon > 0$, pick $n$ such that $ \varepsilon n \ge 1$ and $\forall m \ge n, |a_{m+f(m)} - a_m| \le \varepsilon$. Then, $\forall p \ge m \ge n, |a_p - a_m| \le \sup_{p \ge m} |a_p - a_m| \le \frac 1 m + |a_{m+f(m)} - a_m| \le \frac 1 n + \varepsilon \le 2\varepsilon$. 
A: Let $(n_k)_k$ be an increasing sequence of integers, and let $f(k):=n_k-k>0$. Then $a_{n_k}-a_k\to 0$. In particular, this proves that any two subsequences of $(a_j,j\in\Bbb N)$ have the same limit. 
Now we proves that $a=(a_j,j\in\Bbb N)$ is bounded. Let $(m_k)_k$ be a strictly increasing sequence of integers, and assume that $a$ is not bounded. Then we can find $n_k\uparrow \infty$ such that $|a_{n_k}|\geqslant m_k+1$ for all $k$, hence for $k$ large enough, $|a_k|\geqslant m_k$. This thus give that 
$$\forall (m_k,k\geqslant 1), m_k\uparrow \infty, \quad \liminf_{k\to +\infty}\frac{|a_k|}{m_k}\geqslant 1.$$
This is not possible, taking $m_k:=k+2\max_{1\leqslant j\leqslant n}\lfloor |a_j|\rfloor$.
A: Suppose  $(a_n)$ is not convergent. Then $(a_n)$ is not Cauchy. So, there exists an $\epsilon>0$ such that for each positive integer $k$, there are integers $n_k$ and $m_k$ satisfying $m_k>n_k>k$ and $|a_{n_k}-a_{m_k}|>\epsilon$. One can choose these so that the $n_k$ are distinct.
Now, define $f$ so that for each $k$, $f(n_k)= m_k-n_k$ (give arbitrary values to the other integers). Then $|a_n-a_{n+f(n)}|$ does not converge to zero.
A: I think yes. First of all, the sequence $a_n$ is necessarily bounded above and below. For instance, suppose it were not bounded from above. Then you could select $f$ such that the sequence $a_{n+f(n)}$ diverges to $+\infty$ in such a way as to crush $a_n$. This of course contradicts the hypothesis.
Now as for convergence, we can select a subsequence $a_{n+f(n)}$ that converges to $L=\mathrm{lim~sup}~~a_n$, and thus, using the hypothesis, we get that $a_n$ is convergent.
