Does This Sequence Converge to 2?

[Punkyc7]

use the definition of a sequence to establish the limit
lim([itex]\frac{2n}{n+1}[/itex])=2 Let [itex]\epsilon[/itex]>0, then |[itex]\frac{2n}{n+1}[/itex]-2| <[itex]\epsilon[/itex]. Next we have that | [itex]\frac{2n-2n+-2}{n+1}[/itex]|= |[itex]\frac{-2}{n+1}[/itex]| <[itex]\frac{2}{n}[/itex]. So [itex]\exists[/itex] k[itex]\in[/itex][itex]N[/itex] such that [itex]\frac{2}{k}[/itex]<[itex]\epsilon[/itex]. When n[itex]\geq[/itex]k, we have [itex]\frac{2}{n}[/itex] < [itex]\frac{2}{k}[/itex] <[itex]\epsilon[/itex]. Therefore the limit is 2. BLOCKIs this the right way to do a limit proof?

 

[micromass]

Looks good. Some remarks:

| [itex]\frac{2n-2n+-2}{n+1}[/itex]|

2n+-2 is not something one writes. You'll have to write 2n+(-2) or simply 2n-2. I think you've just made a typo, but I wanted to make sure.

So [itex]\exists[/itex] k[itex]\in[/itex][itex]N[/itex] such that [itex]\frac{2}{k}[/itex]<[itex]\epsilon[/itex].

Why does such a k exist?

 

[Punkyc7]

The Archimedean property right, is that something I should mention? Another question if we weren't given what the limit is equal to do you just make a guess and see if it right?

 

[micromass]

The Archimedean property right, is that something I should mention?

Yes, you should certainly mention that!

Another question if we weren't given what the limit is equal to do you just make a guess and see if it right?

Yes, you need to make an educated guess by looking at the graph or perhaps by using properties of the limit. But I'm sure you will see some easy ways to calculate limits soon.