Proving Limit of Sequence: a1, b1, a2, b2,... Converges to L

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Homework Help Overview

The discussion revolves around proving that the sequence formed by interleaving two convergent sequences, \( a_n \) and \( b_n \), which both converge to a limit \( L \), also converges to \( L \). Participants are exploring the implications of convergence and the application of definitions related to limits.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss defining a new sequence \( c_n \) based on the interleaving of \( a_n \) and \( b_n \). There are attempts to apply the epsilon-delta definition of limits and the pinching theorem. Questions arise regarding the necessity of showing boundedness and the implications of convergence for the new sequence.

Discussion Status

Several participants are actively engaging with the proof structure, with some providing insights into the necessary conditions for convergence. There is a mix of approaches being considered, including the use of specific epsilon values and the relationship between the sequences. No explicit consensus has been reached, but productive lines of reasoning are being explored.

Contextual Notes

Participants are navigating the constraints of formal proof requirements and the definitions of convergence. There is mention of specific epsilon values and the need to establish conditions under which the sequences remain close to \( L \). The discussion reflects a focus on the logical flow of the proof rather than a complete resolution.

matrix_204
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i had this problem in my book that i wasn't able to do. I kinda had the idea of what it involved but just wanted to clear it up with you guys.
So the problem is:
Suppose that an ->L and bn ->L. Show that a1,b1,a2,b2,... converges to L.

So here it seems to me like i can obviously define a new sequence cn, like that in the pinching theorem. But using other definitions(epsilon, etc.) of sequences, how do i come up with a proof of this.
 
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Fix an epsilon. That a_n and b_n converge to L implies there is some N (take each sequence individually, choose an M and M' for each, take the bigger of the two and double that) such that whenever n>N both a_(n/2) and b_(n/2) are within epsilon of L (and therefore c_n is within epsilon of L).
 
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could someone check this proof for me and tell me what is missing as i m not sure if i know anymore:(problem stated above)
Proof:
If a_n ->L and b_n ->L for some L, then for any eps>0 there is a K such that for all n, if n>K, then |a_n - L|<eps/2 and |L - b_n|<eps/2.
Since n>K, |a_n - b_n|</= |a_n - L| + |L - b_n|<eps/2 + eps/2 =eps.
(now to show that it is bounded)
If eps=1, then |a_n - b_n|<1 for some n>K.
This means |a_n - b_K+1|<1 for all n>K.
Thus {a_n:n>K} and {b_n:n>K} are bounded.
So a_n ->L and b_n->L.
-------
 
I don't think you have to show anything is bounded. That will follow automatically once you prove the limit is L. What you want to do is define the sequence by [tex]\{c_n\}_{n\in \mathbb{N}}[/tex] by [tex]c_n=a_{[(n+1)/2]}[/tex] if n is odd and [tex]c_n=b_{(n/2)}[/tex] if n is even. Now fix an epsilon > 0. There are N and M such that whenever n>N, a_(n+1)/2 is within epsilon of L and whenever n>M b_(n/2) is within epsilon of L. Take the bigger of the two, let's say it is P. Then whenever n>P, both a_(n+1)/2 and b_(n/2) are within epsilon of L. That is, whenever n>P, c_n is within epsilon of L.
(sorry for the sloppier version of this above)
 

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