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 &gt; B, then C &gt; 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 &gt; 0, then \gamma &gt; \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 &gt; 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 &lt; ab = \lim a_n(\limsup b_n), and prove that also \gamma &lt; \limsup a_n b_n, i.e., \sup_{n\geq N} a_n b_n &gt; \gamma no matter how large N is. It may help to express \gamma = \alpha\beta where \alpha &lt; \lim a_n and \beta &lt; \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?