Proving the existence of a limit?

pivoxa15

How do you prove the existence but not necessarily the value of a limit?

John Creighto

Why not give an example of such a problem because I think your question is too general.

pivoxa15

Suppose you are given a limit. Show that a limit exists without needing to say what it is.

It is a general question.

John Creighto

Which means there is about a million answers. For instance:
http://en.wikipedia.org/wiki/Absolute_convergence

pivoxa15

Okay, good point. I'll be more specific. I was thinking of limits of sequences only. Or are all limits sequences in which case I haven't narrowed down anything?

John Creighto

Well how about showing the limit of [tex]a_n-a_{n+1}=0[/tex] as n->oo

morphism

This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]

morphism

If you're going to be taking a limit of something, then it had better be a sequence (or something sequence-like). So, no, you haven't narrowed down anything. Do you mean sequences of numbers?

In any case, there are a few existence results that can come in handy. One example is the monotone convergence theorem, which states that if a sequences of real numbers is monotone and bounded, then it must converge.

John Creighto

What if a_n is bound both above and bellow?

pivoxa15

In which case it would be in a compact space. All cauchy sequences in a compact space converge.

morphism

Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.

It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,
[tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]

John Creighto

Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right) [/tex] for all n greater then M

John Creighto

That's a pretty interesting example. Can you find a similar example on the reals?

morphism

What are k and M?

Take the real part of it and see what happens.

pivoxa15

To be more specific, I had a sequence a_m/b_m where a_m and b_m in the limit is 0.

However they are not functions so can't use that rule which I can't spell, l'hospil's rule?

John Creighto

k an M are arbitrary chosen to make the inequality work for a given sequence.

HallsofIvy

Why didn't you say that when you were first asked to be specific?

If all you know is that [itex]a_n[/itex] and [itex]b_n[/itex] go to 0, you can't say anything about whether [itex]a_n/b_n[/itex] converges or diverges. For example, if [itex]a_n= 1/n^2[/itex] and [itex]b_n= 1/n[/itex], it is obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0. And [itex]a_n/b_n= (1/n^2)(n/1)= 1/n[/itex] converges to 0.

But if [itex]b_n= 1/n^2[/itex] and [itex]a_n= 1/n[/itex], it is still obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0 but now [itex]a_n/b_n= (1/n)(n^2/1)= n[/itex] does not converge.