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Limit superior & limit inferior of a sequence

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Fact:
    Let a=lim sup an.
    Then for all ε>0, there exists N such that if n≥N, then an<a+ε

    Theorem 1:
    If lim an = a exists, then lim sup an = lim inf an = a.
    n->∞

    Theorem 2:
    If lim sup an = lim inf an = a, then
    lim an exists and equals a.
    n->∞

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I was trying to see why theorems 1 & 2 are true.
    How can we prove these rigorously?

    I wrote down all the definitions, but still I don't know how to prove theorems 1 and 2.

    Let an be a sequence of real numbers. Then by definition, an->a iff
    for all ε>0, there exists an integer N such that n≥N => |an - a|< ε.

    Also, lim sup an is defined as
    lim sup{an: n≥N}
    N->∞
    (similarly for lim inf)

    Any help is much appreciated! :)
     
  2. jcsd
  3. Jan 15, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For Theorem 2 you also need:
    Let a= lim inf an.
    Then for all ε> 0 there exists N such that if n≥N, then an> a-ε
    If the sequence converges to a, the every subsequence converges to a so the set of all "subsequential limits" is simply {a}.

    Since a= lim sup an and lim inf an, you have both of the above so that, for any ε> 0 there exist N such that if n≥ N then a-ε< an< a+ε.

     
  4. Jan 16, 2010 #3
    1) About theorem 1, I haven't seen the version of the definition that you gave for lim sup and lim inf. My version of the definition is sipmly:
    lim sup an is defined as
    lim [sup{an: n≥N}]
    N->∞

    lim inf an is defined as
    lim [inf{an: n≥N}]
    N->∞

    How can we prove theorem 1 directly by using these definitions (and the definition of "limit")?



    2) Suppose a=lim sup an= lim inf an.
    Then for all ε>0, there exists N1 such that if n≥N1, then an<a+ε
    and for all ε> 0 there exists N2 such that if n≥N2, then an> a-ε
    Should I take N=max{N1,N2}? so that n≥N => an<a+ε and an> a-ε, i.e. |an -a|< ε, and therefore an->a.
    I think in general we should assume that N1 and N2 may be different (i.e. not necessarily the same N). Is this the correct way to prove theorem 2?


    Can somebody help me, please?
    Any help is much appreciated! :)
     
    Last edited: Jan 16, 2010
  5. Jan 16, 2010 #4
    Here are some hints for proving Theorem 1:

    1. I'll assume that you're working over the real numbers, not the extended reals where values of + or - infinity are allowed.

    2. Show (or simply observe, since it's easy) that lim inf [itex]a_n \leq[/itex] lim sup [itex]a_n[/itex].

    3. Show that |[itex]a_n[/itex] - a| < [itex]\epsilon[/itex] implies that [itex]a_n[/itex] < a + [itex]\epsilon[/itex].

    4. What, then, can you say about lim sup [itex]a_n[/itex] in terms of a and [itex]\epsilon[/itex]?

    5. What does the relation in #4 imply about lim sup [itex]a_n[/itex] and a?

    6. Using similar arguments, derive an analogous relation between lim inf [itex]a_n[/itex] and a.

    7. Use the observation in #2 above to finish the proof.

    HTH

    Petek
     
  6. Jan 17, 2010 #5
    Thanks for your hints, but still I don't know how to prove theorem 1 (I have no idea how to prove 2,4,5).

    Here is a proof of theorem 1 from my notes:
    an->a
    => for all ε>0, there exists N such that if n≥N => |an-a|<ε
    So n≥N => a-ε < an < a+ε
    => for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε
    Thus, sup{an: n≥N} -> a as N->∞

    ==============================

    But I really have absolutely no idea why the last two lines (highlighted in blue) are true. What is the point of introducing N' ? And why is it true that for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε?

    Could someone please explain?
     
  7. Jan 18, 2010 #6
    for all n≥N, a-ε < an < a+ε
    => for all N' ≥N, a-ε ≤ sup{an: n ≥ N' } ≤ a+ ε

    Why is this implication true? (particularly the lower bound)

    Help...I am totally confused. Could someone kindly explain?
    Any help is much appreciated! [I'm dying on this proof :( ]
     
  8. Jan 18, 2010 #7
    I'll try to explain this. To simplify the notation, define

    [itex]V_{N'}[/itex] = {[itex]a_n[/itex]: n [itex]\geq[/itex] N'}

    and

    [itex]v_{N'}[/itex] = sup {[itex]V_{N'}[/itex]}

    Then [itex]V_1 \supseteq V_2 \supseteq V_3 \supseteq ...[/itex], and
    [itex]v_1 \geq v_2 \geq v_3 \geq ...[/itex]

    (The second string of inequalities follows from the rule that if S and T are subsets of the reals and S [itex]\supseteq[/itex]T, then sup S [itex]\geq[/itex] sup T, which you should have already covered in your class.)

    Now, using the above notation, we have to show that

    [itex]a - \epsilon \leq v_{N'} \leq a + \epsilon[/itex]

    For the right inequality, we have that [itex]a_n < a + \epsilon[/itex], so [itex]a + \epsilon[/itex] is an upper bound for [itex]V_{N'}[/itex]. Since [itex]v_{N'}[/itex] is the least upper bound for these sets, it follows that [tex]v_{N'} \leq a + \epsilon[/itex], as required.

    To get the left inequality, argue as above, but use inf instead of sup. Define [itex]u_{N'}[/itex] = inf {[itex]a_{N'}: N' \geq N[/itex]}. Show that [itex]a - \epsilon \leq u_{N'}[/itex] similar to the above. Finally, observe that inf [itex]u_{N'} \leq [/itex]sup [itex]v_{N'}[/itex] and you get the left inequality.

    HTH

    Petek
     
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