TITLE: vercongent sequences
QUESTION [12 upvotes]: Definition- We say a sequence $(x_n)$ verconges to $x$ if there exist an $\epsilon>0$ such that for all $N\in \Bbb{N}$, $n\ge N \implies |x_n-x|<\epsilon$.

Loosely speaking, by convergent sequence we mean that a sequence is convergent to some point $x$ if we can confine its infinite "tail" in some neighbourhood of $x$.
In similar wordings, vercongent sequence should mean, that a sequence $(x_n)$ is vercongent to $x$ if  we are given a point in its infinite "tail", say $\mathfrak{n}\in \Bbb{N}$, then we can put a wall around $x$, in which whole tail after $\frak{n}$ must lie.
Question- Give an example of convergent sequence and one that is not vercongent or verdigent
According to me $(x_n)=\{1,\frac12,\frac13, \frac14, \dots \}$ verconges to $0$, also to $1$  as for any given $n\in \Bbb{N}$, all terms after $n$ lies in $\epsilon=\frac{1}{n}$ neighbourhood of $0$, clearly, and for $1$ we  $\epsilon=\frac{n-1}{n}$ works. I guess  a convergent sequence is also vercongent, with the same limit, any many more limits, almost any number works as a limit, am I correct?
But sequence $\{1,-1,1,-1,1,-1,\dots\}$ is not convergent as it is oscillatory but it is vercongent as for any natural number $\frak{n}$, we can always choose $\epsilon=10$ around $0$ or $1$ or $-1$, so it verconges to many numbers, a big enough epsilon will work. Thus vercongent does not imply convergent.
But on the other hand, sequence $\{1,2,3,\dots \}$ neither converges nor verconges.
Please correct me if I am wrong in this understanding of this exercise from Abbott's book "Understanding Analysis."

REPLY [3 votes]: Yes. This horrible distortion of English (and Latin) is a synonym for bounded: a sequence $x_n$ is "vercongent" (to anything you like) iff for some $b$, $|x_n| < b$ for all $n$.