Suppose that the sequence [math] S_{n}[/math] diverges to [math]+\infty[/math].

 

Now if [math]S_{n}[/math] does not diverge to [math]+\infty[/math] , does that imply that [math]S_{n}[/math] goes to a limit s

No. Just imagine a sequence alternating between the values 1 and 0 as a counter example.

No. Just imagine a sequence alternating between the values 1 and 0 as a counter example.

 

You mean, that if the sequence does not diverge to[math]+\infty[/math] then vibrates ( alternates) between two Nos ??

You mean, that if the sequence does not diverge to[math]+\infty[/math] then vibrates ( alternates) between two Nos ??

 

Consider the sequence: 1/n if n is even, 1 - 1/n if n is odd. This doesn't exactly alternate, but it does have two limiting values. You can have arbitrarily many limiting values.

=Uncool-

You mean, that if the sequence does not diverge to[math]+\infty[/math] then vibrates ( alternates) between two Nos ??

I mean that a sequence that alternates between the values 0 and 1 does not "diverge to infinity" and does not converge, either. Therefore, a sequence that does not diverge to infinity but does not converge either exists.

I certainly did not say that a sequence that does not grow arbitrarily and does not converge must always alternate between two numbers. It can obviously alternate between three numbers, too. And it's hopefully obvious that alternating between two or three numbers does not sum up all possibilities, either.

 

EDIT: Could it be that some of your "sequence" is supposed to be "series" or "sum of"? Because on 2nd reading I really wonder whether your opening post "Assume A. Now, if not A ..." perhaps was supposed to make sense?

Edited December 3, 200916 yr by timo

So let us make one point very clear.

 

A sequence or perhaps a series that does not diverge to [math]+\infty[/math] does not converge to a limit.

 

Agree ??

No. A sequence that is constantly equal to 0 does converge to a limit (zero) and does not grow over all bounds (i.e. diverge to infinity). Same goes for the series that results if you add the terms in the sequence up. A sequence that does not grow over all bounds does not have to converge. It still can.

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

 

How can you substantiate that ??

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

How can you substantiate that ??

Let's say I have two sequences that don't diverge to infinity:

• [imath]a_n = (-1)^n[/imath]
  
• [imath]b_n = (\frac{2}{3})^n[/imath]
  

One oscillates, the other converges. QEF.

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

.

 

Sorry,

 

When i asked to substantiate that i mean to prove your claim.

 

An example does not constitute , a proof.

An example does not constitute , a proof.

You made a universal statement. A simple counterexample *is* proof that your claim is false.

You made a universal statement. A simple counterexample *is* proof that your claim is false.

 

What is proof??

What is proof??

 

You claimed that for any sequence that didn't go to infinity, that is, any bounded sequence, that sequence does not converge to any finite limit. the tree showed that there was a sequence that didn't go to infinity that did converge to a finite limit, thereby proving your statement wrong.

=Uncool-

What the hell? I claimed that non divergent sequences can converge or oscillate, for that statement to be true there obviously only needs to be one case of each.

Suppose that the sequence [math] S_{n}[/math] diverges to [math]+\infty[/math].

 

Now if [math]S_{n}[/math] does not diverge to [math]+\infty[/math] , does that imply that [math]S_{n}[/math] goes to a limit s

Let me try to help the confusion here (out of my own empathy -- I always found this subject to be confusing and counter intuitive).

 

The question is asking whether or not S_n's divergence to infinity implies that it goes to a limit. "Implies that" means that we are safe to assume that implication.

 

However, you were given examples of where S_n would *not* diverge to a limit - it can either converge or "fluctuate". Both are AGAINST the suggestion that the claim can imply that Sn goes to a limit.

 

Hence, the answer is no, and it's well shown and demonstrated.

 

If your idea is a full blown mathematical proof, I can't help you there, but I do believe proof by contradiction (hence, showing an example directly contradicting the claim) is, indeed, a proof, even in math.

 

~moo

If your idea is a full blown mathematical proof, I can't help you there, but I do believe proof by contradiction (hence, showing an example directly contradicting the claim) is, indeed, a proof, even in math.

Sure is, reductio ad absurdum. Any proposition that admits a contradiction is non-valid; it must be rejected.

 

triclino appears to still have a problem distinguishing between universal and existential quantifiers. The easiest way to prove an existential quantifier is by example. To prove the proposition some integers are even, all one needs to show is that 2*1=2. Done. [math]a\in S, P(a)\implies \exists x\in S: P(x)[/math]. Proof by construction is a bit tougher, exhaustion even tougher, but they all come down to the same rule, which is existential introduction.

 

Universal and existential quantifiers are related in that the negation of a universal quantifier is an existential quantifier, and the negation of an existential quantifier is a universal quantifier:

 

[math]

\aligned

\neg(\forall x\in S \;P(x)) &= \exists x\in S \;\neg P(x) &&

\text{Negation of a universal quantifier} \\

\neg(\exists x\in S \;P(x)) &= \forall x\in S \;\neg P(x) &&

\text{Negation of an existential quantifier}

\endaligned[/math]

Edited December 5, 200916 yr by D H

Instead of showing your knowledge about quantifiers by using silly examples ,why don't you prove by using quantifiers and their laws ,what is the statement resulting from the negation of the following statement:

 

[math]\forall\epsilon[\epsilon>0\Longrightarrow\exists N(n\in N\wedge\forall n(n\geq N\Longrightarrow S_{n}\geq\epsilon))][/math]

 

Which is the definition of a sequence diverging to[math]+\infty[/math]

 

And thus answering my opening post

Instead of showing your knowledge about quantifiers by using silly examples ,why don't you prove by using quantifiers and their laws ,what is the statement resulting from the negation of the following statement:

 

[math]\forall\epsilon[\epsilon>0\Longrightarrow\exists N(n\in N\wedge\forall n(n\geq N\Longrightarrow S_{n}\geq\epsilon))][/math]

 

Which is the definition of a sequence diverging to[math]+\infty[/math]

 

And thus answering my opening post

Okay, triclino, first off, drop the attitude. People have been participating in this thread trying to help you make sense of it. No one owes you anything, people are here because they attempt to help, and the conversation seem to resort to frustration on both sides. Derailing it further will not help, and it will definitely will not encourage others to help you.

 

Your question was answered in the second post. Since it wasn't clear (admittedly, to me as well), it was answered further in the rest of the thread. Over and over again.

 

You were given a mathematically sufficient proof. You were explained how this proof is sufficient in basic terms. You were given ample examples and math experts explained why the proof by negation is a mathematically accepted proof.

 

If something is still not clear, or, alternatively, if you want to get into detail, I suggest you ask your questions more clearly and with less of an attitude so people can actually help you.

Please, people, don't derail this thread into personal frustrations. Keep it civil.

 

~moo

[math]\forall\epsilon[\epsilon>0\Longrightarrow\exists N(n\in N\wedge\forall n(n\geq N\Longrightarrow S_{n}\geq\epsilon))][/math]

 

Which is the definition of a sequence diverging to[math]+\infty[/math]

You have an error in that expression Ignoring that, you really should not be using [math]\epsilon[/math] here. Mathematicians use epsilon to denote a small number. This is talking about large numbers.

 

In English, a series diverges to infinity if for any positive bound H, no matter how big, there is always some natural number N such that the absolute values of the partial sums S_n exceed this bound H for all n>=N.

 

So what is the negation of this? Simple: The series is bounded.

 

[math]\exists H\in \mathbb R^+:\,\forall n\in \mathbb N \ |S_{n}|<H[/math]

Edited December 6, 200916 yr by D H

 

So what is the negation of this? Simple: The series is bounded.

 

[math]\exists H\in \mathbb R^+:\,\forall n\in \mathbb N \ |S_{n}|<H[/math]

 

Let us stick to sequences

 

The question now is :

 

Does the sequence converge

Some integer n is finite. Is it even or odd?

 

Your question has pretty much the same answer. You have already been given example of a bounded series than converges and another that does not.