TITLE: A variant of Kronecker's approximation theorem?
QUESTION [6 upvotes]: Let $\tau,\sigma\in(0,\infty)$ with $\frac{\tau}{\sigma}\notin\mathbb Q$. By Kronecker's approximation theorem, we know:

(1) For each $x\in \mathbb R$ and $\epsilon>0$, there are $m,n\in\mathbb N$ such that $|x+n\tau-m\sigma|<\epsilon$.

In other words, if you keep adding $\tau$ to $x$, you will eventually come arbitratily close to the set $\sigma\mathbb N$. But what happens if you keep adding values that are just approximately $\tau$?
To make this a precise question, let $(\tau_n)_{n\in\mathbb N_0}\subset (0,\infty)$ with
$$ \tau_{n+1}-\tau_n \xrightarrow{n\to\infty}\tau.$$
Then, the according conjecture is:
(2) For each $x\in \mathbb R$ and $\epsilon>0$, there are $m,n\in\mathbb N$ such that $|x+\tau_n-m\sigma|<\epsilon$.
If one assumes that
$$ \sum_{n=0}^\infty \left((\tau_{n+1}-\tau_n)-\tau \right) \text{ converges in $\mathbb R$,}$$
it is indeed relatively easy to deduce (2) from (1).
QUESTION: If $\sum_{n=0}^\infty \left((\tau_{n+1}-\tau_n)-\tau \right)$ diverges, does (2) still hold?
My ad hoc ideas didn't quite work out and before I start to think deeper about it, I thought I might ask if anyone here knows of any result in this direction.
Thanks a lot in advance!

REPLY [2 votes]: Thank you so very, very much, Ewan Delanoy! Your Lemma is exactly what I needed. Let me say that I'm only posting this as an answer, because it is too long for a comment.
Your Lemma basically says that the number $n$ in (1) can in fact always be taken from the set $\{0,\ldots,M=M(\epsilon)\}$. This is indeed all I need to prove (2) (since this allows to argue in the same manner, as if the series were convergent). Your proof of (2) is a bit off though, as you seem to demand that $\tau_k$ converges to $\tau$, which is not what I assume. For the sake of completeness and clarity, let me rework that part as follows:
Let $\epsilon>0$. Without loss of generality we can assume that $x+\tau_0\ge 0$ and
$$ |(\tau_{k}-\tau_{k-1})-\tau|<\frac{\epsilon}{2M(\epsilon/2)} \quad \text{for all $k\in\mathbb N$.}$$
Hence for all $n\in\{0,\ldots,M(\epsilon/2)\}$ and $m\in\mathbb N$ we have
\begin{align*}
   |x+\tau_n-m\sigma|&=\left|x+\tau_0+\sum_{k=1}^{n}((\tau_{k}-\tau_{k-1})-\tau)+n\tau-m\sigma\right| \\
         &\le \sum_{k=1}^{M(\epsilon/2)}|(\tau_{k}-\tau_{k-1})-\tau| + |x+\tau_0+n\tau-m\sigma| \\
         &\le \frac{\epsilon}{2} + |x+\tau_0+n\tau-m\sigma|.
\end{align*}
Now, thanks to your Lemma, we can choose $n$ and $m$ such that the second summand is also smaller than $\frac{\epsilon}{2}$.
Once again thank you very much for providing me with this essential ingredient (and its neat proof).