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Homework Help: Proof about limit superior of a sequence

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    ra8.JPG

    2. Relevant equations
    N/A

    3. The attempt at a solution
    By definition,
    an->a iff
    for all ε>0, there exists an integer N such that n≥N => |an - 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!
     
  2. jcsd
  3. Jan 21, 2010 #2
    This is a sloppily phrased question.

    You can define [tex]\limsup a_n[/tex] 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 [tex] \sup \{ a_k : k \geq n \}[/tex] converges. Try to figure out what this other definition should be.

    Also, the definition of [tex]\limsup[/tex] given before the question doesn't necessarily apply to the question itself. Your hypotheses don't allow you to conclude that [tex](a_n b_n)[/tex] is a bounded sequence (consider [tex]a_n = 1[/tex] and [tex]b_n = n[/tex]), so you may not be able to compute [tex]\limsup a_n b_n[/tex] 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 [tex]A = B[/tex] by proving that [tex]A \leq B[/tex] and [tex]A \geq B[/tex]. Prove that [tex]A \leq B[/tex] by proving that if [tex]C > B[/tex], then [tex]C > A[/tex]. And so on.
     
  4. Jan 21, 2010 #3
    Thanks. But I really have no idea how to begin...

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

    = (lim an) (lim sup bn)] ??
    n->infinity

    Any more help, please?
     
  5. Jan 21, 2010 #4
    Yes, you can do that.
     
  6. Jan 22, 2010 #5
    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!!!!!
     
  7. Jan 22, 2010 #6
    As I said above, when you want to prove that [tex]A = B[/tex], an efficient way to do it is to prove both [tex]A \leq B[/tex] and [tex]A \geq B[/tex]. 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 [tex]\inf_n 1/n = 0[/tex]. It is obvious that [tex]\inf_n 1/n \geq 0[/tex]: [tex]1/n \geq 0[/tex] for every [tex]n[/tex], which means that [tex]0[/tex] is a lower bound of the set [tex]\{1/n : n \geq 1\}[/tex], and therefore the greatest lower bound of this set must be at least [tex]0[/tex]. To prove the other direction, that [tex]\inf_n 1/n \leq 0[/tex], we show that if [tex]\gamma > 0[/tex], then [tex]\gamma > \inf_n 1/n[/tex]. This is proved by observing that [tex]\gamma[/tex] cannot be a lower bound of [tex]\{1/n : n \geq 1\}[/tex], by the archimedean axiom (choose [tex]n > 1/\gamma[/tex]).

    Use this kind of indirect (comparison of estimates) approach to prove the inequalities in your problem.
     
  8. Jan 26, 2010 #7
    Suppose an->a.
    Let b=lim sup bn.

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

    Take N=max(N1,N2).
    Then n≥N => anbn<(a+ε)(b+ε)
    => sup{anbn: 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 anbn ≥ 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)> ???

    May someone help me, please?
     
    Last edited: Jan 26, 2010
  9. Jan 26, 2010 #8
    Choose [tex]\gamma < ab = \lim a_n(\limsup b_n)[/tex], and prove that also [tex]\gamma < \limsup a_n b_n[/tex], i.e., [tex]\sup_{n\geq N} a_n b_n > \gamma[/tex] no matter how large [tex]N[/tex] is. It may help to express [tex]\gamma = \alpha\beta[/tex] where [tex]\alpha < \lim a_n[/tex] and [tex]\beta < \limsup b_n[/tex].

    Note that you need to deal with the special case where one or more of your limits is [tex]0[/tex], as you can't underestimate [tex]0[/tex] by nonnegative numbers.
     
  10. Jan 26, 2010 #9
    hmm...sorry, I don't understand. Can you explain more, please?

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

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

    Thanks!
     
  11. Jan 26, 2010 #10
    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?
     
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