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Homework Help: Lim Sup, Lim Inf Proof

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    prove lim inf Sn= - Lim Sup (-Sn)

    2. Relevant equations

    3. The attempt at a solution

    I know how to run the proof when dealing with the sup and inf. But the limsup and liminf is another story. I wanna do it the same way by saying that limsup (-SN) >= Sn but i know I cannot say that. From what I gather all I can say that the Sups >= Limsup. But that does not help me relate it to the sequence itself. Need a push to get me going.
  2. jcsd
  3. Dec 7, 2009 #2
    Sn is a sequence of numbers, right?
    So remember, liminf (Sn)= lim n->infinity (inf{Sm: m>=n})
    Now, you know that inf{Sm: m>=n} = -sup{-Sm: m>=n}.
    That's my push
  4. Dec 7, 2009 #3
    I need more of a nudge

    I am with you on that inf(Sn : n> N} = -sup{-Sn: n >= N} Though I do not think I can say lim as n goes to infinity and finish it off. I know as n goes to invintiy that lim sup <= -sup{-Sn: n >= N} and inf{Sm: m>=n} <= Lim inf. I was thinking this would run similar to the inf Sn = -sup Sn. So I would have to prove -sup Sn is a lowerbound and also the greatest lower bound.
  5. Dec 8, 2009 #4
    I did some reading and slepted on it so how about the following

    Let S0 be the lim sup of -Sn. Thus as n -> infinity -Sn <= S0. By the order postulates it follows that -S0 <= Sn making, -S0 a lowerbound for Sn as n -> infinity. Now assume L >= -S0, but L <= Sn, making L also a lowerbound. By order postulates -L >= -Sn as n-> infinity making L an upperbound. By defintion of the suprenum, S0 <= -L but by the order postualtes L<= -S0 which is a contradiction -S0 = lim inf of Sn.

    Is the basic idea for the lim sup and lim inf is that it forms a box where the limit is allowed to be, and as n ---> infinty Sn <= lim Sup and same logic for the lim inf. Also the infs are an increasing sequence and sups are decreasing and essential squeeze the box to get the limit value if it exists?

    Thanks in advance
  6. Dec 8, 2009 #5
    yeah, you're right in your description of the basic idea of liminf and limsup.

    Why not?

    That's not completely rigorous. Note that it could be the case that for all n, -Sn>S0. For example, let Sn=-1/n, so Sn=1/n. Then S0 = limsup -Sn = lim -Sn = 0, even though for all n, -Sn=1/n>0.

    So you really do need to talk about limits.
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