Proof about limit superior of a sequence

[kingwinner]

Homework Statement

Homework Equations

N/A

The Attempt at a Solution

By definition,
a_n->a iff
for all ε>0, there exists an integer N such that n≥N => |a_n - a|< ε

But other than stating and looking at the definitions, I really have no clue how to proceed on this problem.

Any help is very much appreciated!

 

[ystael]

This is a sloppily phrased question.

You can define $$\limsup a_n$$ without using the limit operator. In fact, the fact that you can use the limit operator is a theorem, i.e., it needs to be proved that the sequence $$\sup \{ a_k : k \geq n \}$$ converges. Try to figure out what this other definition should be.

Also, the definition of $$\limsup$$ given before the question doesn't necessarily apply to the question itself. Your hypotheses don't allow you to conclude that $$(a_n b_n)$$ is a bounded sequence (consider $$a_n = 1$$ and $$b_n = n$$), so you may not be able to compute $$\limsup a_n b_n$$ using this definition.

Once you correct the definition and the question to get past all that, try a strategy that often works for proofs involving limits, suprema, and infima: Prove that $$A = B$$ by proving that $$A \leq B$$ and $$A \geq B$$. Prove that $$A \leq B$$ by proving that if $$C > B$$, then $$C > A$$. And so on.

 

[kingwinner]

Thanks. But I really have no idea how to begin...

For the right hand side, can I say that lim sup b_n is a constant, so I can pull it out of the limit?
i.e. lim [a_n(lim sup b_n)]
n->infinity

= (lim a_n) (lim sup b_n)] ??
n->infinity

Any more help, please?

 

[ystael]

Yes, you can do that.

 

[kingwinner]

hmm...but what should I do next? no idea where to begin...

Could someone please give me more hints/guidelines/steps of the proof?

thanks!

 

[ystael]

As I said above, when you want to prove that $$A = B$$, an efficient way to do it is to prove both $$A \leq B$$ and $$A \geq B$$. And when you are dealing with a quantity like a limit or a supremum, which is usually described in terms of estimates, you want to carry this one step further, proving the inequality itself indirectly.

Let me give an example, and prove that $$\inf_n 1/n = 0$$. It is obvious that $$\inf_n 1/n \geq 0$$: $$1/n \geq 0$$ for every $$n$$, which means that $$0$$ is a lower bound of the set $$\{1/n : n \geq 1\}$$, and therefore the greatest lower bound of this set must be at least $$0$$. To prove the other direction, that $$\inf_n 1/n \leq 0$$, we show that if $$\gamma > 0$$, then $$\gamma > \inf_n 1/n$$. This is proved by observing that $$\gamma$$ cannot be a lower bound of $$\{1/n : n \geq 1\}$$, by the archimedean axiom (choose $$n > 1/\gamma$$).

Use this kind of indirect (comparison of estimates) approach to prove the inequalities in your problem.

 

[kingwinner]

Suppose a_n->a.
Let b=lim sup b_n.

First I'm trying to prove that limsup a_nb_n ≤ ab.
Given ε>0.
a_n->a => there exists N₁ s.t. if n≥N₁, then a_n<a+ε.
b=lim sup b_n => there exists N₂ s.t. if n≥N₂, then b_n<b+ε

Take N=max(N₁,N₂).
Then n≥N => a_nb_n<(a+ε)(b+ε)
=> sup{a_nb_n: n≥N}≤(a+ε)(b+ε)
Then taking limits, I got the result because the inequality holds for ALL epsilon>0, so we can get rid of the epsilons.

How can I prove the other direction? i.e. limsup a_nb_n ≥ ab (I am trying to modify my proof but I got stuck)
Given ε>0.
a(n)->a => there exists N1 s.t. if n≥N1, then a(n)>a-ε. <-----this is true for sure
b=limsup b(n) => there exists N2 s.t. if n≥N2, then b(n)> ?[/color]

May someone help me, please?

 

[ystael]

Choose $$\gamma < ab = \lim a_n(\limsup b_n)$$, and prove that also $$\gamma < \limsup a_n b_n$$, i.e., $$\sup_{n\geq N} a_n b_n > \gamma$$ no matter how large $$N$$ is. It may help to express $$\gamma = \alpha\beta$$ where $$\alpha < \lim a_n$$ and $$\beta < \limsup b_n$$.

Note that you need to deal with the special case where one or more of your limits is $$0$$, as you can't underestimate $$0$$ by nonnegative numbers.

 

[kingwinner]

hmm...sorry, I don't understand. Can you explain more, please?

So we're trying to prove that limsup a_nb_n ≥ ab

Is it possible to mimic/modify my proof above, but instead of finding upper bounds, we find lower bounds for a_n and b_n? How can we find lower bound for b_n that will get the proof to work?

Thanks!

 

[kingwinner]

Here is my try...

Given ε>0.
a(n)->a => there exists N1 s.t. if n≥N1, then a(n)>a-e.
b=limsup b(n) => there are infinitely many n's s.t.then b(n)>b-e

Case 1: (a,b>0)
So for infinitely many n's, we have a(n)b(n)>(a-e)(b-e) for ALL e>0.
=> for infinitely many n's (not necessaily of the form n≥N), we have a(n)b(n)≥ ab.

But now I DON'T see how we can use this to prove that sup{a(n)b(n): n≥N} ≥ab for some N (and hence our result limsup a(n)b(n) ≥ ab).

Can somebody help, please?