# $\epsilon-N$ definition of convergent sequence: Assumptions in order for $N=1$

The formal definition of a convergent sequence as stated in wiki is the following.

For each real number $$\epsilon>0$$, there exists $$N\in \mathbb N$$ such that for any $$n\geq N$$, we have \begin{align*} \vert x_n-x\vert <\epsilon \;\; (*). \end{align*}

Now, if we consider a positive sequence $$\epsilon_n$$ instead of $$\epsilon>0$$ in $$(*)$$, I wonder if there are any sufficient assumptions on the sequence $$\epsilon_n$$ in order to obtain $$N=1$$. More precisely, I would like to have this:

For any positive sequence (+some more assumptions) $$\epsilon_n>0$$, we have \begin{align*} \vert x_n-x\vert <\epsilon_n \;\; \text{for all } n\geq 1. \end{align*}

Would this be possible in the first place? My guess is that since in the usual $$\epsilon-N$$ definition, the value of $$N$$ depends in principle on $$\epsilon$$, then in my case I would need perhaps an assumption on $$\epsilon_1$$. However, the ideas are really vague at the moment.

I hope my question is clear. Any help or hint will be much appreciated. Thanks in advance!

• Only constant sequences converge in this sense. Commented Nov 15, 2021 at 12:32
• @KaviRamaMurthy So there are no assumptions that could be imposed in $\epsilon_n$ in order to achieve this convergence for non-constant sequences? Commented Nov 15, 2021 at 12:34

Let $$\epsilon > 0$$ and $$\epsilon_n = |x_n - x| + \epsilon$$. Then $$|x_n - x| = \epsilon_n - \epsilon < \epsilon_n$$.
This is because convergence, by definition, is only concerned with what happens towards infinity. A series can be arbitrarily chaotic in the first $$N$$ steps and still be convergent.