# Homework Help: "super convergence" proof

1. Oct 20, 2014

### neemer

1. The problem statement, all variables and given/known data

Suppose ($a_n$) is a sequence and let $l\in\mathbb R$. Let us say that ($a_n$) is "super convergent" to $l$ if there exists $N\in\mathbb N$ such that for every ε>0 we have $n \geq N$ ⇒ |($a_n - l$|<ε . Show that if ($a_n$) super converges to l then ($a_n$) converges to $l$ in the usual sense. What about the converse?

2. Relevant equations
The usual definition of convergence as given in my textbook is:

Let ($a_n$) be a sequence and let $l\in\mathbb R$. Then ($a_n$) converges to $l$ if for every ε>0 there exists $N(ε)\in\mathbb N$ such that $n \geq N(ε)$ ⇒ |($a_n - l$|<ε

3. The attempt at a solution

The only difference I see between the super convergent definition and usual definition is that for super convergence N does not depend on ε as it does in the usual definition. Therefore no matter how small an ε is chosen, any $n \geq N$ will imply ($a_n - l$|<ε . So even if $N$ is dependant on ε as it is in the usual definition, any N will work so the dependence does not matter. Therefore super convergence implies convergence.

The second part about the converse is what I am confused about. I'm pretty sure that for converse I'm supposed to show that if ($a_n$) converges in the usual sense, that it also must "super converge" which is obviously not true because if N is dependent on ε for regular convergence, a random N cannot be selected to have the super convergence statement hold. I'm pretty sure I'm on the right track with this question I just don't know how i would go about showing this last part.

2. Oct 20, 2014

### Dick

Yes, the converse is obviously not true. If you think about what 'super convergence' means, it means the sequence must be constant for n>N. There are sequences that converge that are not eventually constant. Use that for a counterexample. That proves the converse is not true.

3. Oct 21, 2014

### neemer

Makes sense thanks.

4. Oct 21, 2014

### PeroK

You should note and try to prove the following:

If $\forall \ \epsilon > 0, \ \ |x| < \epsilon$ then $x = 0$