# Limits of Subsequences

1. Feb 26, 2008

### soul

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

2. Feb 26, 2008

### CompuChip

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. Feb 27, 2008

### sutupidmath

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. Feb 28, 2008

### HallsofIvy

Staff Emeritus
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 $\epsilon$= (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 $2\epsilon$ of each other, they cannot both be within $\epsilon$ of L, a contradiction.

5. Oct 14, 2010

### vertciel

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. Oct 14, 2010

### Design

I was wondering how when you choose the max of the two, how you just add the two parts of the sub sequences?

7. Oct 17, 2010

### Mr.Miyagi

Suppose $$\lim a_n=L$$. Let $$a_{n_i}$$ be an arbitrary subsequence. We wish to prove that $$\lim a_{n_i}=L$$.

We know that $$\forall \epsilon>0$$ $$\exists N$$ such that $$n\geq N$$ implies $$|a_n-L|<\epsilon$$.

We need to show that $$\forall \epsilon>0$$ $$\exists I$$ such that $$i\geq I$$ implies $$|a_{n_i}-L|<\epsilon$$. But if we choose $$I$$ such that $$i\geq I$$ implies that $$n_i\geq N$$, then $$i\geq I$$ implies $$|a_{n_i}-L|<\epsilon$$. The existence of such an $$I$$ 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.

 Just noticed the date of the OP's post... oh well.

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