Proving the Convergence of Sequence a_n to a: How to Show Unique Partial Limits?

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The discussion focuses on proving that the sequence a_n, defined as (x1, y1, x2, y2, ...), converges to a given that both sequences x_n and y_n converge to the same limit a. It is established that if a_n has a unique partial limit, it converges to that limit. The participants clarify that by taking the maximum of the indices N1 and N2 for the converging sequences, one can show that a_n also converges to a. This approach confirms that for sufficiently large n, the distance |a_n - a| can be made less than any given ε. The conclusion is that the sequence a_n indeed converges to the limit a.
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i have that lim x_n=lim y_n=a
and we have the sequence (x1,y1,x2,y2,...)
i need to show that this sequence (let's call it a_n) converges to a.

well in order to prove it i know that if a_n has a unique partial limit then it converges to it to it, but how do i show here that it has a unique partial limit?
i mean if we take the subsequences in the even places or the odd places then obviously those subsequences converges to the same a, but i need to show this is true for every subsequence, how to do it?
thanks in advance.
 
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loop quantum gravity said:
i have that lim x_n=lim y_n=a
and we have the sequence (x1,y1,x2,y2,...)
i need to show that this sequence (let's call it a_n) converges to a.

well in order to prove it i know that if a_n has a unique partial limit then it converges to it to it, but how do i show here that it has a unique partial limit?
i mean if we take the subsequences in the even places or the odd places then obviously those subsequences converges to the same a, but i need to show this is true for every subsequence, how to do it?
thanks in advance.
I think you are making too much of this. Since xn converges to a, given \epsilon> 0 there exist N1 such that if n> N1 then |x_n-a|< \epsilon. Since yn converges to a, given \epsilon> 0 there exist N2 such that if n> N1 then |y_n-a|< \epsilon.

What happens if you take N= max(N1, N2)?
 
yes i see your point.
if we take the max of the indexes then from there, we have that either way a_n equals x_n or y_n, and thus also a_n converges to a, cause from the maximum of the indexes both the inequalities are applied and thus also |a_n-a|<e, right?
 
Yes, that is correct.
 
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