A variant of Kronecker's approximation theorem? 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!
 A: Your claim is true, and here is why.
Summary of proof. A compactness argument allows one to use a strengthened version of Kronecker's theorem that is "more uniform" in $x$  namely :
Main lemma. There is a constant $M$ (depending only on $\sigma,\tau$ and $\epsilon$ and not on $x$)  such that for any $x \geq 0$, there are integers $(n,m)\in[0,M]\times {\mathbb N}$ with $|x+n\tau-m\sigma| \lt \epsilon$.
Detailed proof.
Replacing $(x,\tau,\sigma,\epsilon)$ with $(\frac{x}{\sigma},\frac{\tau}{\sigma},1,\frac{\epsilon}{\sigma})$, we may assume without loss that $\sigma=1$.
For $n,m\in {\mathbb N}$, let
$$A_{n,m}= \bigg\lbrace X\in {\mathbb R} \bigg| |X+n\tau-m| \lt\epsilon\bigg\rbrace.\tag{3}$$
Then, Kronecker's usual theorem says that whenever $\tau$ is irrational, there are nonnegative integers $n(x),m(x)$ with $x\in A_{n(x),m(x)}$.
Then $\bigcup_{x\in [0,1]} A_{n(x),m(x)}$ is an open covering of $[0,1]$. Since $[0,1]$ is compact, there is a finite subset $I\subseteq [0,1]$ such that $\bigcup_{x\in I} A_{n(x),m(x)}$ is still a covering of $[0,1]$. Denote by $M$ the maximum value of $n(x)$ or $m(x)$ when $x$ varies in the finite set $I$. We have then that
$$
[0,1] \subseteq \bigcup_{0 \leq n,m \leq M}
A_{n,m}. \tag{4}
$$
(4) means that for any $x\in [0,1]$, we can find $n,m$ with $0 \leq n,m \leq M$ such that $$(*) : \quad |x+n\tau-m| \leq \epsilon.$$
Now, if $x\geq 1$, and we put $x'=x-\lfloor x \rfloor$ (the fractional part of $x$), then $x'\in [0,1]$ so that $|x'+n'\tau-m'| \leq \epsilon$ for some $(n',m')=(n(x'),m(x'))$. But then ($*$) holds also for $(n',m'+\lfloor x \rfloor)$ in place of $(n,m)$. We deduce that
$$
 {\mathbb R}^+ \subseteq \bigcup_{0 \leq n \leq M, m\geq 0}
A_{n,m}. \tag{4'}
$$
This concludes the proof of the main lemma. Let us now prove (2). Using $\frac{\epsilon}{2}$ instead of $\epsilon$ in the main lemma, there is a $M>0$ such that for any $y \geq 0$, there are integers $(n(y),m(y))\in[0,M]\times {\mathbb N}$ with
$$|y+n(y)\tau-m(y)| \lt \frac{\epsilon}{2}.\tag{5}$$
Let $\delta >0$ be a positive constant whose value is to be decided later. By hypothesis, there is a $k_0$ such that $x+\tau_1+\sum_{k=1}^{k_0-1}\tau_{k+1}-\tau_k \geq 0$
and $|\tau_{k+1}-\tau_k-\tau| \leq \delta$ for any $k\geq k_0$.
Let $y=x+\tau_1+\sum_{k=1}^{k_0-1}\tau_{k+1}-\tau_k=x+\tau_{k_0}$ ; we know that $y$ is nonnegative. By (5),
$$\bigg|x+\tau_{k_0}+n(y)\tau-m(y)\bigg| \lt \frac{\epsilon}{2}.\tag{6}$$
On the other hand, we have
$$ \bigg| \sum_{k=k_0}^{k_0+n(y)-1} \tau_{k+1}-\tau_k-\tau \bigg| \leq n(y)\delta \leq \delta M. \tag{7}$$
Adding (6) and (7) and using the triangle inequality, we obtain
$$
\bigg|x+\tau_{k_0+n(y)}-m(y)\bigg|=\bigg|x+\sum_{k=1}^{k_0+n(y)-1}(\tau_{k+1}-\tau_k)-m(y)\bigg| \lt \frac{\epsilon}{2}+\delta M.
$$
Taking $\delta=\frac{\epsilon}{2M}$, we are done.
A: 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).  
