Convergent sequences coming back from infinity This may seem like a dumb question, but is it possibly for a sequence to jump to infinity and after some term later we can find an $N$ such that $n > N \implies |a_n - \ell| < \epsilon$? Because if it is, doesn't that mean convergent sequences does not necessary have to be bounded? 
Or is the problem that once it jumps to infinity, it is immediately divergent and there is no hope of finding an $N$ that gives $|a_n - \ell| <\epsilon$?
 A: A convergent sequence does necessarily have a bounded tail, if you will. There is nothing wrong with having "unbounded" terms at the (relative) start of your sequence. Some might cringe at the idea of setting term no. 3 to equal $\infty$, though, as infinity is not a number. Rather, in some manner of speaking it's a concept of "unboundedness".
Of course there are exceptions, the easiest one being the augmented real number line $\Bbb R \cup \{-\infty, +\infty\}$. But then you will find arithmetic to be difficult to define in a meaningfull way, so you need to tread carefully when you use it in conjunction with, for instance, the operation $+$. I'd recommend you read up on "Hilbert's hotel" for a nice thought experiment on how just how difficult it is to make sense of common operations when $\infty$ is involved.
Edit: Seing your last comment, there is no such "point of no return" on this side of infinity. Either it's a real number, with a (more or less) comprehensible size, or it's not. Either way, convergence only cares about the last part of your sequence, be that from the xkcd-numbered term on or even later.
A: If you generalize from sequences to transfinite sequences, then yes, you can. Let
$$x_\beta = \begin{cases}
\beta &\text{if $\beta < \omega$} \\
5 &\text{if $\beta \ge \omega$.}
\end{cases}$$
Then this sequence "goes to infinity" (in a sense) and jumps back down to $5$.
