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!

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

1 Answer 1


This is an unsatisfying answer, but ...

Let $\epsilon > 0$ and $\epsilon_n = |x_n - x| + \epsilon$. Then $|x_n - x| = \epsilon_n - \epsilon < \epsilon_n$.

So, technically, yes there are sufficient assumptions. You would have to be more specific what kind of assumptions you need. In general I doubt that you would find any assumptions that are stricter than the one given above, or in any way useful.

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.


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