Point belongs to the boundary - real analysis

[Dassinia]

Hello,
I have some trouble to solve this exercise

Homework Statement

E={ (-1)ⁿ (8n+7)/(4n-1) : n ∈ℕ}
Show that 2∈[PLAIN]http://www.ilemaths.net/img/smb-bleu/derivepartielle.gifE

Homework Equations

The Attempt at a Solution

We have to show that (2-r,2+r)∩ E ≠∅ and (2-r,2+r)∩ ℝ/E ≠∅
If I take the part where (-1)ⁿ=1 i will note it E''
We have
(8n+7)/(4n-1)=2+9/(4n-1)
I want to show that there is a point y so that y∈(2-r,2+r)∩ E'' i.e N :
2-r<2+9/(4N-1)
The thing is that we also have 2+r<2+9/(4N-1) but we are supposed to have 2+9/(4N-1) between 2-r and 2+r and then show that N ∃ by Archimedes' Principle, so what is wrong ?

Thanks

 

[vela]

I want to show that there is a point y so that y∈(2-r,2+r)∩ E'' i.e N :
2-r<2+9/(4N-1)
The thing is that we also have 2+r<2+9/(4N-1) but we are supposed to have 2+9/(4N-1) between 2-r and 2+r and then show that N ∃ by Archimedes' Principle, so what is wrong?

Why do you think $2+r < 2+\frac{9}{4N-1}$? You get to choose N so that $2+\frac{9}{4N-1} < 2+r$ holds.

 

[Dassinia]

Thank you !