vercongent sequences

Claim: Any real-valued sequence is “vercongent” if and only if it is bounded, in which case it “verconges” to every real number.

Proof of sufficiency: If $(x_n)_{n\in\mathbb N}$ is bounded, then there exists some $M>0$ such that $|x_n|<M$ for all $n\in\mathbb N$. Hence, the sequence “verconges” to $0$. In fact, it “verconges” to any number $x\in\mathbb R$, as $|x_n-x|\leq|x_n|+|x|<M+|x|$ for any $n\in\mathbb N$.

Proof of necessity: Suppose that $(x_n)_{n\in\mathbb N}$ “verconges” to $x\in\mathbb R$ and let $\varepsilon >0$ be the constant of “vercongence.” Then, letting $N=1$ in the definition of “vercongence,” one has that for any $n\in\mathbb N$, $$|x_n|\leq |x_n-x|+|x|<\varepsilon+|x|.$$

Added (generalization to metric spaces): Let $(X,d)$ be a metric space.

Definition: A sequence $(x_n)_{n\in\mathbb N}$ in $X$ is vercongent to $x\in X$ if there exists some $\varepsilon>0$ such that $d(x_n,x)<\varepsilon$ for all $n\in\mathbb N$.

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Claim: Any sequence in $X$ is vercongent if and only if it is bounded, in which case it verconges to every $x\in X$.

Proof of sufficiency: If $(x_n)_{n\in\mathbb N}$ is bounded, then the set $\{x_n\,|\,n\in\mathbb N\}$ can be included in a sufficiently large open ball. That is, there exists some $M>0$ and $y\in X$ such that $d(x_n,y)<M$ for all $n\in\mathbb N$. In fact, the sequence is vercongent to any $x\in X$, since $$d(x_n,x)\leq d(x_n,y)+d(y,x)<M+d(x,y)\quad\text{for every $n\in\mathbb N$}.$$

Proof of necessity: Suppose that $(x_n)_{n\in\mathbb N}$ verconges to $x\in X$ and let $\varepsilon >0$ be the constant of vercongence. Then, it is easy to check that $$\{x_n\,|\,n\in\mathbb N\}\subseteq B(\varepsilon, x).$$

An alternative, and perhaps more widely used, definition of boundedness of a subset $A\subseteq X$ of a metric space $(X,d)$ is that its diameter $$\operatorname{diam}(A)\equiv\sup_{x,y\in A}d(x,y)\tag{$\clubsuit$}$$ is finite. To see that this is equivalent to the definition used above, suppose first that there is some $z\in X$ and $M>0$ such that $A\subseteq B(M,z)$. Then, for any $x,y\in A$, one has $$d(x,y)\leq d(x,z)+d(z,y)<2M,$$ so that $\operatorname{diam}(A)\leq2M<\infty$. Conversely, if the supremum in ($\clubsuit$) is finite, then pick any $x\in A$ (if $A$ is empty, it is trivially included in any open ball). Then, for any $y\in A$, $$d(y,x)\leq\operatorname{diam}(A)<\operatorname{diam}(A)+1,$$ so that $$A\subseteq B(\operatorname{diam}(A)+1,x).$$

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 $\mathbb{R}$)?