What does this definition of a sequence mean? I came across this problem in a book of mixed mathematical excersices:
A sequence $a_1,a_2,...$ we denote with $(a_n)$. A sequence $(a_n)$ of positive numbers $a_n$ is said to be a "zero-sequence" if there for every $ε > 0$ exist an integer $n_ε \geq 1$ such that $a_n \lt ε$ for every $n \geq n_ε$. Let $(a_n)$ and $(b_n)$ be sequences with positive integers and let $c_n = a_n + b_n$. Show that $(c_n)$ is a "zero-sequence" if and only if both $(a_n)$ and $(b_n)$ are "zero-sequences".
Can someone please explain with words what the definition of a "zero-sequence" means and why it is interesting to know wether a sequence is a "zero-sequence", i.e. what unique properties do  "zero-sequences" have? You may assume that I have no previous knowledge of what this type of sequences are about and that I am not familiar with sequences in general.
 A: The question is a bit open. However, here is one possible explanation of convergences with words:
"A sequence converges to zero, if every $\varepsilon$-ball around $0$ contains all elements of the sequence with just finitely many exceptiones."
(With $\varepsilon$-ball, I refer to the interval $(-\varepsilon,\varepsilon)$. In higher dimensions this would be replaces by the ball of all points which have distance less than $\varepsilon$ to $0$.)
How does this relate to your definition?


*

*First we descide which $\varepsilon$ value we choose. (or in my terms, which $\varepsilon$-ball around zero.) 
However, it is important to make sure that everything we do holds for all possible choices we do here.

*Saying that only finitely many exceptions are allowed, can be translated to
"there is one last outliner, after which all other elements of the sequence remain in the $\varepsilon$-ball". Let us use a number $n_{\varepsilon}-1$ to denote the indices of that last outliner. 
Notice that the value $n_{\varepsilon}$ depends on the choice of $\varepsilon$.


*Saying that all other elements after the last outliner are inside the $\varepsilon$-ball, is nothing else then
$$
a_n \in (-\varepsilon,\varepsilon), \quad \text{for all } n \geq n_{\varepsilon}
$$
Since your numbers of the sequence are positive, can replace this by
$$
a_n < \varepsilon, \quad \text{for all } n \geq n_{\varepsilon}
$$
Why do we care for zero-sequences?


*

*If $a_n$ denotes the error of an approximation, where $n$ is the level of details of our simulation. Than it is quite natural to ask, if we somehow get arbitrary close to $0$ with out error.
Here $\varepsilon$ can be seen as the largest error we tolerate. And $n_{\varepsilon}$ is the number after which we stay within the error tolerance. Convergence plays a big role in numerics and applied math.

*However, there are arbitrary many applications within and outside of mathematics.
Convergences allows us to approximate. It allows us to find new numbers (like real numbers as the limit of rational numbers). We can show that equations have solutions with convergences, ...
The list of applications is endless. Convergences is at the very core of mathematical analysis.


*

*Zero convergences is related to convergence to any value, via
$a_n - C$ is a zero sequence, if and only if $a_n$ convergences to $C$.

