Divergent sequence

[Darío]

Hi forum!

 

I have that the sequence [latex]a_n=\{2-(-1)^n\}[/latex] not converges. I must show this with the rigorous definition.

 

I think use [latex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\ge N}:|a_n-\ell|\ge\epsilon[/latex]

 

How i can continue?

[DrRocket]

Hi forum!

 

I have that the sequence [latex]a_n=\{2-(-1)^n\}[/latex] not converges. I must show this with the rigorous definition.

 

I think use [latex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\ge N}:|a_n-\ell|\ge\epsilon[/latex]

 

How i can continue?

 

First you need to review the definition of convergence. Divergence is a failure to converge.

 

The condition that you state would imply divergence but is somewhat stronger (assuming that you could prove this for any [math] \ell[/math], which you will not be able to do). This sequence has two cluster points, but no limit.

 

You might try writing out the first few terms of your sequence. The pattern that emerges ought to suggest to you a reason why the sequence fails to converge.

[Darío]

I know that the sequence can write as

 

[latex]a_n=\begin{cases} 1\ if\ n\ is\ even \\ 3\ if\ n\ is\ odd \end{cases}[/latex]

 

So the sequence is osillating, therefore not exist the limit.

 

But i need prove this formally, then i need the definition.

 

how i can the rigorous definition?

[DrRocket]

I know that the sequence can write as

 

[latex]a_n=\begin{cases} 1\ if\ n\ is\ even \\ 3\ if\ n\ is\ odd \end{cases}[/latex]

 

So the sequence is osillating, therefore not exist the limit.

 

But i need prove this formally, then i need the definition.

 

how i can the rigorous definition?

 

 

Start by writing out the definition of a convergent sequence.

 

Ask yourself if it is possible for a convergent sequencve to have two distinct cluster points. Prove your answer.

[Darío]

This is equivalent to prove that the limit is unique, something trivial... Thanks!

[triclino]

This is equivalent to prove that the limit is unique, something trivial... Thanks!

 

This is the wrong approach definitely.

 

The sequence has two sub sequences each having a limit. But the sequence itself has no limit.

 

So the uniqueness of limit approach will not work.

 

The proof of non existence of the limit for this particular sequence using the 'ε'definition is very difficult to prove indeed.

 

I wonder if anyone in this forum can do it.

 

But let us put things in their correct prospective .

 

First of all the correct definition for convergence is the following:

 

[math]x_{n}[/math] converges in R if and only if................. [math]\exists a (a\in R\wedge\forall\epsilon(\epsilon>0\Longrightarrow\exists k(k\in N\wedge\forall n(n\geq k\Longrightarrow |x_{n}-a|<\epsilon)))[/math]

 

 

 

Meaning : There exists an aεR ,such for all ε>0 there exists a kεN such that for all [math]n\geq k[/math] [math]|x_{n}-a|<\epsilon[/math]

 

Now the negation of the above statement ( which b.t.w is also very difficult to prove) is the following:

 

 

 

[math]\forall a(a\in R\Longrightarrow\exists\epsilon(\epsilon>0\wedge\forall k(k\in N\Longrightarrow\exists n(n\geq k\wedge |x_{n}-a|\geq\epsilon))))[/math]

 

Meaning: Given any real No a we must be able to find an ε>0 such that for any Natural No k we have to find a natural No n ,[math] n\geq k [/math] such that :[math] |x_{n}-a|\geq\epsilon[/math].

 

At a first glance this is very difficult to assimilate ,let alone to ,to prove .

 

 

However this is analysis .

 

Where do we start ??

 

As we usually do with nearly all limits we start from the end of the statements (this is a trial and error process) and work our way opposite to proof that we have to write later.

 

I.e in our case let us start with [math]|x_{n}-a|\geq\epsilon[/math] inequality.

 

We observe that for n even the inequality becomes : [math] |1-a|\geq\epsilon[/math], and

 

with n odd the inequality becomes : [math]|3-a|\geq\epsilon[/math].

 

Well,half of the work is done.

 

Can you carry on from here ...................................???

[DrRocket]

This is the wrong approach definitely.

 

The sequence has two sub sequences each having a limit. But the sequence itself has no limit.

 

These sentences are contradictory. If a sequence has two subsequences that converge to different limits that in and of itself is enough to show that the original sequence does not converge.

 

So, not only is it a valid approach, you have inadvertently made the essential observation but failed to understand the significance,.

 

Not only is non-convergence not "very difficult to prove", it is trivial with this observation. Perhaps you should gain an understanding of the suubject before offering bogus advice.

Edited December 18, 2011 by DrRocket

[triclino]

Hi forum!

 

I have that the sequence [latex]a_n=\{2-(-1)^n\}[/latex] not converges. I must show this with the rigorous definition.

 

I think use [latex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\ge N}:|a_n-\ell|\ge\epsilon[/latex]

 

How i can continue?

 

He asked for the 'ε' approach ,did he not???

 

 

 

The condition that you state would imply divergence but is somewhat stronger (assuming that you could prove this for any [math] \ell[/math], which you will not be able to do). This sequence has two cluster points, but no limit.

 

You might try writing out the first few terms of your sequence. The pattern that emerges ought to suggest to you a reason why the sequence fails to converge.

 

 

This is equivalent to prove that the limit is unique, something trivial... Thanks!

 

This is the wrong approach definitely.

 

The sequence has two sub sequences each having a limit. But the sequence itself has no limit.

 

So the uniqueness of limit approach will not work.

 

The proof of non existence of the limit for this particular sequence using the 'ε'definition is very difficult to prove indeed.

 

I wonder if anyone in this forum can do it.

 

 

These sentences are contradictory. If a sequence has two subsequences that converge to different limits that in and of itself is enough to show that the original sequence does not converge.

 

So, not only is it a valid approach, you have inadvertently made the essential observation but failed to understand the significance,.

 

Not only is non-convergence not "very difficult to prove", it is trivial with this observation. Perhaps you should gain an understanding of the suubject before offering bogus advice.

 

What is the matter,apart from not knowing how to solve the problem the way that the OP asked you have problems with the language??

Edited December 18, 2011 by triclino

[DrRocket]

He asked for the 'ε' approach ,did he not???

What is the matter,apart from not knowing how to solve the problem the way that the OP asked you have problems with the language??

 

No, he asked to solve the problem using the definition. If you understood mathematics, which you clearly do not, then it would be quite obvious how to use "epsilons" to prove that no convergent sequence can have more than one cluster point. This result remains true in any metric space and indeed in any Hausdorff topological space. Using epsilons to show that the real numbers constitute a Hausdorff space is the heart of the matter, though that terminilogy is not usually introduced in an elementary calculus class. But the gist of the idea is that any two distimct points lie in disjooint small open intervals and no sequence can be eventually in both of them, though it can be in both frequently.

 

I, in fact, did provide him guidance toward a very valid approach to solve the problem -- using the definition. You very clearly failed to recognize that fact and also failed to recognize the nature of the problem and the natural (and very easy) solution.

 

Apparently the language and mathematics difficulties lie entirely with you.

 

Please refrain from usiing all capital letters to emphasize stupidit comments. They are quite easy to recognize without the added emphasis.

[triclino]

No, he asked to solve the problem using the definition. If you understood mathematics, which you clearly do not, then it would be quite obvious how to use "epsilons" to prove that no convergent sequence can have more than one cluster point. This result remains true in any metric space and indeed in any Hausdorff topological space. Using epsilons to show that the real numbers constitute a Hausdorff space is the heart of the matter, though that terminilogy is not usually introduced in an elementary calculus class. But the gist of the idea is that any two distimct points lie in disjooint small open intervals and no sequence can be eventually in both of them, though it can be in both frequently.

 

I, in fact, did provide him guidance toward a very valid approach to solve the problem -- using the definition. You very clearly failed to recognize that fact and also failed to recognize the nature of the problem and the natural (and very easy) solution.

 

Apparently the language and mathematics difficulties lie entirely with you.

 

Please refrain from usiing all capital letters to emphasize stupidit comments. They are quite easy to recognize without the added emphasis.

 

 

Will the above change the very fact that you do not know how to prove that the sequence does not converge ,by using the negation of the definition of convergence ??.

No ,definitely no

 

And the funny thing is that,as usually, i get red flags from the forum,approving your pompous ignorance and rudeness

 

But i do not care .You know why?? Because that strengthens my belief :

 

That nobody in this forum can give a proof the way that the OP asked

Edited December 19, 2011 by triclino

[DrRocket]

Will the above change the very fact that you do not know how to prove that the sequence does not converge ,by using the negation of the definition of convergence ??.

No ,definitely no

 

And the funny thing is that,as usually, i get red flags from the forum,approving your pompous ignorance and rudeness

 

But i do not care .You know why?? Because that strengthens my belief :

 

That nobody in this forum can give a proof the way that the OP asked

 

 

Wrong.

 

Once you make the obvious observation that the sequence has more than one cluster point, you have proved -- using the definition -- that the sequence fails to converge.

 

The fact that any two distinct points are contained in disjoint intervals (which you can formalize using epsilons if you like bhy takikng epsilon to be one third of the distance separating the points) is all that is needed.

 

The fact that you fail to recognize that fact speaks volumes.

[Cap'n Refsmmat]

Cut it out, guys. If someone has made an error, correct it, rather than insulting them.