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

In summary, the conversation is about proving that if a sequence converges to L and another sequence also converges to L, then a combined sequence of alternating terms from the two sequences also converges to L. The proof involves defining a new sequence and showing that it is bounded and within a given epsilon of L. One person suggests that the proof may be missing something and the other person clarifies the steps involved.
  • #1
matrix_204
101
0
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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  • #2
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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  • #3
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.
-------
 
  • #4
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)
 

What is a limit of a sequence?

A limit of a sequence is a value that the terms of the sequence approach as the index of the terms increases without bound.

What does it mean for a sequence to converge to a limit?

If a sequence has a limit, it means that as the index of the terms increases, the terms of the sequence get closer and closer to the limit value. In other words, the difference between the terms and the limit becomes smaller and smaller.

How do you prove that a sequence converges to a limit?

To prove that a sequence converges to a limit, you must show that for any given epsilon (ε) greater than 0, there exists a positive integer N such that for all n greater than or equal to N, the absolute value of the difference between the term and the limit is less than ε.

What is the epsilon-N definition of a limit of a sequence?

The epsilon-N definition of a limit of a sequence states that if a sequence converges to a limit L, then for any given epsilon (ε) greater than 0, there exists a positive integer N such that for all n greater than or equal to N, the absolute value of the difference between the term and the limit is less than ε.

Can a sequence have more than one limit?

No, a sequence can only have one limit. If the terms of a sequence approach more than one value as the index increases without bound, then the sequence does not have a limit.

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