Troubles with number sequences So consider a sequence $\{a_n = \dfrac{1}{n-2000}, n \geq 1\}$.
Despite the problem with $n = 2000$, it seems to be approaching to 0, since
$\forall \varepsilon > 0: \exists N = [2000+ \dfrac{1}{\varepsilon}]: \forall n > N: |\dfrac{1}{n-2000}| < \varepsilon$ 
where $[x]$ denotes a floor function (an integer part of $x$) 
However we know that every convergent sequence is bounded 
Sequence is bounded, when $\exists C > 0: \forall n \geq 1: |a_n| \leq C$. But what can be done with element $a_{2000}$? 
Most of the problems in books don't include those troubles (like with division by 0), but anyways I have some concerns about it
 
Perhaps, it's pretty illegal to consider the sequence above. However, if we really aren't allowed, then I've got another misunderstanding. When we have two convergent sequences $\{a_n\}, \{b_n\}$, then $\{ \dfrac{a_n}{b_n}\}$ also converges, IF $\displaystyle \lim_{n \to \infty} b_n \neq 0$ (states wiki or some of the real analysis sources). Proving that, we can, of course, see, that $b_n \neq 0, \forall n > N$, starting from some number. But what if there exists some element that $b_n = 0$, when $n \leq N$? I mean why didn't some sources include condition $b_n \neq 0, \forall n \geq 1$?

I hope you could understand the question :c
 A: For your first question, the sequence $a_n=\frac{1}{n-2000}$ is undefined for $n=2000$. However, in practice what is usually meant by such formulas is that you should consider the sequence for $n>2000$. Almost every theorem about sequences remains true if you replace "for all $n\in\mathbb{N}$" with "for all $n$ sufficiently large" i.e. "for all $n\geq n_0$ for some $n_0\in\mathbb{N}$". In that sense you will see that it is common to see people refer to $a_n=\frac{1}{n-2000}$ as bounded, even though in a strict sense we shouldn't actually allow it as a sequence (starting from $n=1$) in the first place.
The correct thing to say is that $\left\{\frac{1}{n-2000}\right\}_{n=2001}^{\infty}$ is a bounded sequence, but often people sweep such issues under the rug and just say that $a_n$ is bounded for short, with the understanding that what is really meant is that it is bounded for those indices where it is defined. Usually this won't cause any confusion, but sometimes it can, so it's good you noticed this.
For your second question, similarly, if $a_n\to A, b_n\to B\ne 0$, then as you say we know there is some $N$ such that for all $n\geq N$ we have $b_n\ne 0$, and so starting from this $N$ it makes sense to talk of the sequence $\frac{a_n}{b_n}$. You are right that this sequence is not necessarily defined for all $n$, but since in the definition of the limit we only really care about large $n$, so it is fine to say that $\frac{a_n}{b_n}\to\frac{A}{B}$ even though strictly speaking the sequence on the left side is only defined from some point onward and to be completely rigorous we should write $\left\{\frac{a_n}{b_n}\right\}_{n=N}^{\infty}\to\frac{A}{B}$.
One way to make sure we stay rigorous while not complicating our arguments would be to define $\lim_{n\to\infty} a_n=L$ to make sense for any sequence that is defined from some $N$ onwards, and one may say that this is already implicit in the standard limit definition, because in "there exists $N$ such that for all $n\geq N$ ..." this $N$ can in particular be chosen to be after the last index where $a_n$ is undefined, if there are such indices.
To conclude, these sort of inaccuracies are common, and it's good that you noticed them, but the point is that it doesn't really matter, which is why these things tend to be swept under the rug: always being precise about them would over complicate otherwise simple statements, and what's important is that we know what the true meanings of the statements are and know how to write them completely rigorously if the need ever arises.
Practically speaking, if I was given this sequence $a_n=\frac{1}{n-2000}$ in an assignment in a calculus course I would write this at the beginning of the solution: "notice that $a_n$ is undefined for $n=2000$ so it is understood that we should look instead at the sequence $\{a_n\}_{n=2001}^{\infty}$", but if it was an advanced course I wouldn't even bother writing that (unless it became pertinent to be explicit about this during the solution).
