State a diverging sequence $a_x$ such that $\lim_{x\to\infty}\left[a_{f(x)}-a_x\right]=0$ 
Is there a non-converging sequence $a_n$ such that 
   for all strictly increasing $f:\mathbb{N}\longrightarrow\mathbb{N}$
$$\lim_{n\to\infty}\left[a_{f(n)}-a_n\right]=0\,?$$

Initially I thought $a_n=\log(n)$ might work, but it fails for $f(n)=kn$, $k\in \mathbb{N}, k\geq 2$.
I'm not sure how to go from here. Maybe we can use mod?
 A: In this case there is no such example because:
$(a_{n})$ diverges $\Leftrightarrow \exists \epsilon>0, \forall N \in \mathbb{N}, \exists (p_{N} > q_{N} \geq N) \in \mathbb{N}^{2}, |a_{p_{N}}-a_{q_{N}}|>\epsilon$, i.e. if and only if $(a_{n})$ is not a Cauchy sequence.
You can pick $p_{N}, q_{N}$ such that $f(q_{N}) = p_{N}$ allows you to define a strictly increasing function from $\mathbb{N} \rightarrow \mathbb{N}$ (in order to prove this, you will need to use the fact that you can make the difference $p_{N}-q_{N}$ as large as you wish - if this were not true, you can prove that $(a_{n})$ would be Cauchy). 
Now that's a contradiction since our stated limit cannot be $0$ with such a definition of $f$...
A: Let the sequence $a_n = n$ and $f(n) = n$.
Your sequence is divergent since $n$ goes to infinity. The series $a_n - f(n)$, however, is convergent since it is $0$ for any $n$ you pick
A: There is no such sequence. Suppose for example $(a_n)$ is bounded and we're in the oscillating case. Then $a_n$ has at least two distinct subsequential limits, say $L_1,L_2\in \mathbb R.$ Let $f(n)$ strictly increase to $\infty$ along indices giving a limit of $L_1.$ Then $a_{f(n)} - a_n$ does not coverge, as it is the difference of a convergent sequence and a divergent sequence. This is just one case, but it should show the way in handling other cases.
