# Need help with upper limit of sequence.

1. Nov 8, 2012

### c0dy

1. The problem statement, all variables and given/known data
Prove that,
$$s^{*} = \lim_{n \rightarrow \infty} \sup_{k \geq n} s_k$$
Assume that $s^{*}$ is finite.

2. Relevant equations
Definition of $s^{*}$ is here: http://i.imgur.com/AWfOW.png

3. The attempt at a solution
I started out writing what I know.
By assuming $s^{*}$ is finite, then $\{s_k\}$ is bounded above so a supremum exists.
I'm unclear what exactly $$\sup_{k \geq n} s_k$$ means. Fixing n and finding supremum of {s_k} for k >= n and then letting n -> oo? I would think if there is an upper limit for {s_k} and for all n < k, as n ->oo then {s_n} will converge to that upper limit. And I have a feeling the Theorem 3.17 in the image might be applicable to this problem?

2. Nov 10, 2012

### c0dy

Any hints?

3. Nov 10, 2012

### Zondrina

$\displaystyle\sup_{k ≥ n} s_k$ means that you want to take the supremum of the set sk generated by all numbers which are greater than or equal to your n.

4. Nov 10, 2012

### c0dy

Ok that makes sense. Then from there I could say:

Let $E = \{s_k\} ^{\infty}_{k=n}$. Since $s^*$ is finite, then $E$ is bounded from above, $E \subset \{s_k\}$ and $E$ is not empty, then a supremum exists in $E$. And then taking limit as $n \rightarrow \infty$, $E$ would consist of only sup$E$ which is exactly $s^*$