Let $(x_n)_{n \in \mathbb{N}}, x_n\in \mathbb{R^+}$ be a increasing diverging sequence.
Let $\epsilon(n) : \mathbb{N} \to \mathbb{R^+_*}$ be a decreasing strictly positive function.
I also add the condition that the $x_n$ are "spaced out".
$x_{n+1} \geq x_n+\epsilon(n)$
$(1)$ Does there always exist an arithmetic progression of the form $p_n = n\alpha$ with $\alpha \in \mathbb{R}$ such as for an infinite number of $n \in \mathbb{N}$, $|x_n-p_n| \leq \epsilon(n)$.
For example if I take $x_n = n + \frac{1}{n}$, and $\epsilon(n) = \frac{1}{2^n}$. I don't see why the progression $p_n = \pi n$ could not go arbitrarily close to $x_n$ an infinite amount of times.
This stems from an exercice I was given to prove which is related to the question.
The exercice states
Let $f:\mathbb{R^+}\to\mathbb{R^+}$ be a continuous function.
Show that
$(2)$ $(\forall \alpha \in \mathbb{R},\space \alpha > 0 \land \lim \limits_{n \in \mathbb{N} \space n \to +\infty} f(\alpha n) \to 0) \implies \lim \limits_{x \in \mathbb{R} \space x \to +\infty} f(x) \to 0$
It is easy to show that this is true with uniform continuity, however for a simple continuous function I have tried to find a counterexample to give myself some intuition about why that should be true.
Let $g_{t, w}(x):\mathbb{R^+}\to\mathbb{R^+}$ be a triangle of width $w$ and height $1$ centered around $t$.
$g_{t,w}(x) = max(0,1-\frac{|x-t|}{w})$
Let $G:\mathbb{R^+}\to\mathbb{R^+}$ $G(x) = \sum \limits_{n=0} \limits^{+\infty} g_{x_n,\epsilon(n)}(x)$
Since the $x_n$ are spaced out G is continuous as a sum of continuous functions with disjoint support.
If $(1)$ is false then I can find some $G$ of this form which will satisfy the first part of the implication of $(2)$ but clearly does not converges to 0. Therefore $$ \space \space \space \space \space \lnot (1) \implies \lnot (2) \\ \Leftrightarrow (2) \implies (1) $$
But since I was asked to prove $(2)$ it should be true, therefore (1) should be true too but it looks weird since $\epsilon(n)$ can get very small very fast.