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Limits of Subsequences

by soul
Tags: limits, subsequences
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soul
#1
Feb26-08, 01:22 PM
P: 62
If two subsequences of a sequence {an} have different limits, does {an} converge? and Why?Could you prove it?
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CompuChip
#2
Feb26-08, 01:24 PM
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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.
sutupidmath
#3
Feb27-08, 07:05 PM
P: 1,633
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!

HallsofIvy
#4
Feb28-08, 06:51 AM
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PF Gold
P: 39,553
Limits of Subsequences

Specifically, you can do this: suppose the subsequence of {an} has subsequence {an}i which converges to P and subsequence {an}j which converges to Q.

Assume that {an} converges to L, and take [itex]\epsilon[/itex]= (1/2)|P- Q|. For any N, there will be n1> N such that an1, in the first subsequence, is arbitratily close to P and n2> N such that an2, 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.
vertciel
#5
Oct14-10, 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:

Design
#6
Oct14-10, 09:21 PM
P: 62
Quote Quote by vertciel View Post
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:

I was wondering how when you choose the max of the two, how you just add the two parts of the sub sequences?
Mr.Miyagi
#7
Oct17-10, 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_n-L|<\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 contra-positive 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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