# Thread: show that sup S =1

1. ## show that sup S =1

show that sup S =1 where S = $\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \}$
proof:
1) $\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S$
2) if v< 1 want to prove $\exists s \in S$ such that $v
$1-v>0 \to \exists N^2 \ such \ that \ \frac {1}{N^2} < 1-v$ (by using Archimedean property ) "this step true or not ??"
$v < 1- \frac {1}{N^2} , \ \forall \ N \ is \ even$

2. Originally Posted by flower3
show that sup S =1 where S = $\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \}$
proof:
1) $\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S$
2) if v< 1 want to prove $\exists s \in S$ such that $v
$1-v>0 \to \exists N^2 \ such \ that \ \frac {1}{N^2} < 1-v$ (by using Archimedean property ) "this step true or not ??"
$v < 1- \frac {1}{N^2} , \ \forall \ N \ is \ even$

The first step is wrong: take odd n's and the limit is -1, and take now even n's and the limit is 1 ==> the limit doesn't exist.
And precisely this gives you the solution: take all the elements of the sequence with even n and show their limit is 1. Now just mention that for odd n's you have negative terms...

Tonio

3. Originally Posted by tonio
The first step is wrong: take odd n's and the limit is -1, and take now even n's and the limit is 1 ==> the limit doesn't exist.
And precisely this gives you the solution: take all the elements of the sequence with even n and show their limit is 1. Now just mention that for odd n's you have negative terms...

Tonio
i really don't know any thing about "limits" so i solve this Q by def of sup !!!!

4. Originally Posted by flower3
i really don't know any thing about "limits" so i solve this Q by def of sup !!!!
Then what does the " $\to$" in $
\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S
$
mean?

5. Originally Posted by HallsofIvy
Then what does the " $\to$" in $
\{ (-1)^{n} - \frac {1}{n^2} ; n \in N \} \to 1 \geq s , \forall s \in S
$
mean?
means "implies"