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Limits of Subsequences 
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#1
Feb2608, 01:22 PM

P: 62

If two subsequences of a sequence {an} have different limits, does {an} converge? and Why?Could you prove it?



#2
Feb2608, 01:24 PM

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P: 4,300

By definition, we say that a (sub)sequence converges if its limit exists (and is finite).
By definition, if the limit exists, it is unique. You may prove this for yourself. For example, the sequence {1, 1, 1, 1, 1, 1, ....} has two converging subsequences {1, 1, 1, 1, ...} and {1, 1, 1, 1, ...} (actually, there are infinitely many, as long as it ends in just 1's or just 1's) which converge to 1 and 1 respectively. The sequence itself has no limit though. 


#3
Feb2708, 07:05 PM

P: 1,635

there is a theorem i guess, which states that if {an} converges then every subsequence of it converges and that to the same nr as {an}. Hence if we can find at least two subsequences of a sequence {an} that converge do different nrs, that is have different limits, then the sequence {an} does not converge!



#4
Feb2808, 06:51 AM

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PF Gold
P: 39,348

Limits of Subsequences
Specifically, you can do this: suppose the subsequence of {a_{n}} has subsequence {a_{n}}_{i} which converges to P and subsequence {a_{n}}_{j} which converges to Q.
Assume that {a_{n}} converges to L, and take [itex]\epsilon[/itex]= (1/2)P Q. For any N, there will be n1> N such that a_{n1}, in the first subsequence, is arbitratily close to P and n2> N such that a_{n2}, in the second subsequence is arbitrarily close to Q. If they are not within [itex]2\epsilon[/itex] of each other, they cannot both be within [itex]\epsilon[/itex] of L, a contradiction. 


#5
Oct1410, 08:27 PM

P: 63

Hello everyone,
I have tried to write a proof based on HallsofIvy's response, posted below. However, I am not able to derive a contradiction from what I have at the moment. Could someone please assist me with the conclusion of this proof? Thank you very much. Attempt: 


#6
Oct1410, 09:21 PM

P: 62




#7
Oct1710, 10:43 PM

P: 47

Suppose [tex]\lim a_n=L[/tex]. Let [tex]a_{n_i}[/tex] be an arbitrary subsequence. We wish to prove that [tex]\lim a_{n_i}=L[/tex].
We know that [tex]\forall \epsilon>0[/tex] [tex] \exists N[/tex] such that [tex]n\geq N[/tex] implies [tex]a_nL<\epsilon[/tex]. We need to show that [tex]\forall \epsilon>0[/tex] [tex] \exists I[/tex] such that [tex]i\geq I[/tex] implies [tex]a_{n_i}L<\epsilon[/tex]. But if we choose [tex]I[/tex] such that [tex]i\geq I[/tex] implies that [tex]n_i\geq N[/tex], then [tex]i\geq I[/tex] implies [tex]a_{n_i}L<\epsilon[/tex]. The existence of such an [tex]I[/tex] is guaranteed by the definition of a subsequence (something to check). So what this proof says is that if a sequence is convergent, then all it's subsequences converge to the same limit. Now take the contrapositive of this statement and compare it with your question. I hope that helped. [edit] Just noticed the date of the OP's post... oh well. 


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