What is the proof for the limit superior?

[NihalRi]

Homework Statement

2. Relevant equation
Below is the definition of the limit superior

The Attempt at a Solution

I tried to start by considering two cases, case 1 in which the sequence does not converge and case 2 in which the sequence converges and got stuck with the second case.

I know intuitively that there exists a K such that Mk < a is the second case but I can not think of how to show this. I attempted to consider cases again, like monotone increasing, decreasing, or even use the definition of couchy sequence but was not getting anywhere. Is there a way to reach the same conclusion in my second case or is my approach for this proof completely off? I would gladly supply any additional information and would greatly appreciate any help.

 

[andrewkirk]

You don't need to consider cases and it doesn't matter whether the sequence converges.

Let $u=(s+a)/2$. Can you show that for large enough $n$, the amount $M_n$ must be less than $u$? What does that tell us about the relationship between $x_n$ and $a$?

 

[NihalRi]

You don't need to consider cases and it doesn't matter whether the sequence converges.

Let $u=(s+a)/2$. Can you show that for large enough $n$, the amount $M_n$ must be less than $u$? What does that tell us about the relationship between $x_n$ and $a$?

for a large enough n, wouldn't $M_n$ = s?
a > s
a + s > s + s = 2s
(a+s)/2 > s
so s < u

I'm still trying to see the next part

 

[andrewkirk]

for a large enough n, wouldn't $M_n$ = s?

Not necessarily. But it will be less than u. Why is that? (Consider the definition of lim sup)

 

[member 587159]

It might be useful to observe that

$$\limsup a_n = \inf_{n=1}^\infty \{\sup_{k=n}^\infty a_k\}$$

 

[member 587159]

I have more time now. Using my last formula, if $a > \limsup a_n := L$, then there is $n \geq 1$ such that $\sup_{k \geq n} a_k < a$. But then $a_k \leq \sup_{k \geq n} a_k < a$ for all $k \geq n$, which was what we had to prove.

It remains to show that my last formula is correct. This is however easy, since $\sup_{k\geq n} a_k$ is a non-increasing sequence in $n$ and thus by the monotonuous convergence theorem, the sequence of suprema converges to its infinum.