How Do You Prove a Sequence Converges Below a Certain Limit?

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Homework Help Overview

The discussion revolves around proving a property of convergent sequences, specifically addressing the condition where the limit of a sequence is less than a certain value. Participants explore the implications of this condition and how it relates to the behavior of the sequence.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the limit being less than a certain value and question the correctness of interpretations regarding the relationship between the limit and the sequence terms. Some suggest using the definition of convergence to establish the proof, while others provide counterexamples to challenge assumptions.

Discussion Status

The discussion is active, with participants offering insights and corrections. Some have provided potential approaches to the proof, while others have raised questions about the assumptions made in the original post. There is a mix of interpretations being explored, and guidance has been offered regarding the definition of convergence.

Contextual Notes

Participants note the importance of correctly interpreting the conditions of the problem, particularly the relationship between the limit and the value p. There is an acknowledgment of the need for clarity in the definitions being used.

kreil
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I have this problem on my HW assignment and I just need a little help getting started...


Let {an} be a convergent sequence and let [itex]l=\lim_{n{\rightarrow}\infty}a_n[/itex]

Prove that if l < p then:


[tex]\exists N{\in}N \forall n \ge N : a_n < p[/tex]


Thanks!
 
Last edited:
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You have something backwards I think. If L > P then intuitively you expect all the [tex]a_n[/tex] to eventually be close to L and thus greater (not less) than P, right? Can you make progress with this intuitive picture in mind?
 
kreil said:
...
Let {an} be a convergent sequence and let [itex]l=\lim_{n{\rightarrow}\infty}a_n[/itex]
Prove that if l > p then:
[tex]\exists N{\in}N \forall n \ge N : a_n < p[/tex]
...
Counterexample: Let an = 1 - (1/(n+1)). Then [itex]\lim_{n\rightarrow\infty}a_n = 1[/itex]. Choose p = 1/10 < 1. Then [itex]\forall N \in \mathbb{N} \exists n\ge N : a_n > p[/itex].
 
Physics Monkey said:
You have something backwards I think. If L > P then intuitively you expect all the [tex]a_n[/tex] to eventually be close to L and thus greater (not less) than P, right?
Not necessarily. Let an = 1/(n+1). Then [itex]\lim_{n\rightarrow\infty} a_n = 0[/itex]. If you choose p=9/10 > 0, all the an's near l are less than p. I think a qualifier is missing somewhere.
 
No, hypermorphism, you missed the condition in kreil's original post that L > P. In your case L < P from which it follows, as you have indicated, the [tex]a_n[/tex] will eventually be less than P and near L.
 
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Whoops. :smile: That's probably the right question, then. In that case, kreil should appeal to the definition of convergence.
 
sorry its l < p

kreil
 
I did some work on it and I'm pretty sure I got it...

Show: an<p

l < p [itex]\implies[/itex] p-l > 0

[tex]l=\lim_{n{\rightarrow}\infty}a_n < p <br /> <br /> \iff \forall \epsilon > 0 \exists N \in N \forall n \ge N: |a_n-l| < \epsilon[/tex]

Choose e = p - l > 0

[tex]\exists N \in N \forall n \ge N: |a_n-l|<\epsilon=p-l[/tex]

and thus

[tex]\exists N \in N \forall n \ge N: a_n<l+{\epsilon}=l+p-l=p[/tex]



Josh
 
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kreil said:
sorry its l < p

Yeah, great, that makes more sense. Also, your proof looks good.
 

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