Question：Is description
$$\forall\varepsilon\gt0,\exists N\in\mathbb{Z^+},\text{s.t.}\forall n\gt N,\left| a_n-a_N\right|\lt\varepsilon$$
equivalent to convergence of the suquence $\{a_n\}$?
Attempt:
$(\Leftarrow)$Suppose $\{a_n\}$ is convergent.
Then for $\varepsilon/2\gt0$, $\exists N_0\in\mathbb{Z^+},\text{s.t.}\forall n\ge N_0,\left| a_n-A\right|\lt\varepsilon/2$.
Select $N=N_0$, then we have
$$\forall n\gt N,\left|a_n-a_N\right|\lt\left|a_n-A\right|+\left|a_N-A\right|\lt\varepsilon$$
Obstacles:
I have no way to prove the sufficiency or to give any counterexamples to falsify.
If your sequence fulfills what you wrote, then for any $\;\epsilon>0\;\exists N\in\Bbb N\; $ s.t. if $\;n, m\in\Bbb N\;,\;\;n,m>N\;$ we get:
$$|a_n-a_m|=|a_n-a_N+a_N-a_m|\le|a_n-a_N|+|a_m-a_N|<2\epsilon$$
and thus your sequence is Cauchy and then in any complete metric space, e.g. in the reals $\;\Bbb R\;$ or in the complex $\;\Bbb C\;$ , the sequence converges.