Real analysis: Limit superior proof

In summary, the conversation discusses the definition of a limit of a sequence of real numbers, and a theorem regarding the limit of the supremum of the sequence. The theorem states that for any given ε>0, there exists an integer N such that if n≥N, then all terms in the sequence will eventually be less than a+ε. The conversation also includes a proof of the theorem and a request for help in completing the proof.
  • #1
kingwinner
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Homework Statement


Definition:
Let (an) be a sequence of real numbers. Then we define
lim [sup{an: n≥k}] = lim sup an
k->∞

(note: sup{an: n≥k} = sup{ak,ak+1,ak+2,...} = bk
(bk) is itself a sequence of real numbers, indexed by k)

Theorem:
Let a=lim sup an.
Then for all ε>0, there exists N such that if n≥N, then an<a+ε

Homework Equations


N/A

The Attempt at a Solution


I am trying to prove the theorem.
Proof:
Let bk=sup{an: n≥k}.
By the definition of "a" as a limit of supremums (bk->a), we have that
for all ε>0, there exists an integer N such that if k≥N, then
|sup{an: n≥k} - a| = |bk -a|< ε
=> sup{an: n≥k} < a + ε
By definition of an upper bound, sup{an: n≥k} ≥ an if n≥k.
So the above shows that if k≥N and n≥k, then an< a+ ε.

Now I am stuck...how should I continue? (I need to prove that: for all ε>0, there exists N such that if n≥N, then an<a+ε. But from my work so far, I'm really feeling hopeless)

There are so many different subscripts (e.g. n and k) to keep track of that I am really puzzled and frustrated now.

Can someone please help me out and show me the correct way to prove the theorem?
Any help is very much appreciated!
 
Last edited:
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  • #2




Thank you for your post. I can see that you have made a good start in your proof of the theorem. Let me try to guide you towards the next steps.

First, let's recall what we are trying to prove: for any given ε>0, we want to show that there exists an integer N such that if n≥N, then an<a+ε. In other words, we want to show that eventually, all terms in the sequence an will be less than a+ε.

Now, you have already shown that for any given ε>0, there exists an integer N such that if k≥N, then sup{an: n≥k} < a+ε. But this is not quite what we want, because the supremum of the terms in the sequence an is not necessarily the same as the terms themselves.

So, how can we use the fact that the supremum is less than a+ε to show that the terms in the sequence an are also less than a+ε? Think about what the supremum represents - it is the smallest upper bound for the terms in the sequence an. Can you see how this can help you prove the theorem?

I hope this helps you make progress in your proof. If you are still stuck, feel free to reply to this post and I will try to guide you further. Good luck!
 

FAQ: Real analysis: Limit superior proof

What is the limit superior in real analysis?

The limit superior, also known as the upper limit or limsup, is a concept in real analysis that represents the largest limit point of a sequence. It is denoted by lim sup n→∞ xn and is defined as the supremum of all the subsequential limits of the sequence.

Why is the limit superior important in real analysis?

The limit superior is an important concept in real analysis because it helps us to determine whether a sequence has a limit or not. It also provides information about the behavior of a sequence at infinity and is used in many proofs and theorems in analysis.

How do you prove the limit superior in real analysis?

The most common way to prove the limit superior in real analysis is by using the epsilon-delta definition. This involves showing that for any given epsilon, there exists a corresponding delta such that the difference between the limit superior and the value of the sequence is less than epsilon for all n greater than delta.

What is the difference between limit superior and limit inferior?

The limit superior and limit inferior are two different concepts in real analysis. While the limit superior represents the largest limit point of a sequence, the limit inferior, also known as the lower limit or liminf, represents the smallest limit point. In other words, the limit superior is the supremum of all the subsequential limits, while the limit inferior is the infimum.

Can a sequence have both a limit superior and limit inferior?

Yes, a sequence can have both a limit superior and limit inferior. If the two values are equal, then the sequence is said to have a limit. If the limit superior is greater than the limit inferior, then the sequence is said to be divergent. However, if the two values are different, then the sequence is said to be oscillating and does not have a limit.

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