vercongent sequences 
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."

You're correct that vercongent is more general than convergent, but note that vercongent does not actually have to do with the tail of $(x_n)$. In fact, since for all $N\in\mathbb{N}$ we have $n\geq N\implies |x_n-x|<\epsilon$, we can take $N=1$ in particular to see that $|x_n-x|<\epsilon$ for all $n\in\mathbb{N}$. Thus, a vercongent sequence is bounded; can you prove that every bounded sequence is vercongent (to any point in $\mathbb{R}$)?
