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Homework Help: Lim Sup

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    find the limsup and liminf of [tex]a_{n}=(-1)^{n}\left( 1+\frac{1}{n} \right)[/tex]


    2. Relevant equations
    Given a sequence of real numbers a_n, the supremum limit (also called the limit superior or upper limit), written lim sup is the limit of
    [tex]a_{n}=sup_{(k>=n)}a_{k}[/tex]


    3. The attempt at a solution
    I started by using the identity sup(f+g) = sup(f) + sup(g) to turn the initial equation into a sum after multiplying it out. Then I separated the problem into two different limits. Now it seems obvious that the limsup of the given equation is 1 but I'm not sure how to prove that result.

    limsup[tex](-1)^{n}[/tex] is clearly one. I showed this by setting n=2k, for some natural number k, and then saying for some n > k, [tex](-1)^{n}=1[/tex]. Then the sup is 1, and the limsup is 1 also.

    I'm running into problems finding the limsup[tex]\frac{(-1)^{n}}{n}[/tex].
    Do I have to use the same n = 2k, because the limit here goes to 0, leaving me with the result I want, but I'm not sure that: a) how to go about writing this out and b) if I'm even proceeding correctly.

    I think if I can work through the limsup portion of the problem, it will be easy to apply the reasoning to the liminf portion.

    Thanks.
     
  2. jcsd
  3. Mar 18, 2010 #2

    Gib Z

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    Homework Helper

    I'm not sure what you did with setting n=2k, I can't see what the logical argument is exactly, could you please explain it?

    Also, a good method to approach the problem could be to use the Squeeze law.

    May I also note that It's great to see someone who has given their problem a substantial amount of help. I'm not saying other people around here don't do that as well, but there are far too many who give no effort at all. To everyone read, a good attempt will make us want to help you. Good work Edellaine.
     
  4. Mar 18, 2010 #3

    HallsofIvy

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    Science Advisor

    For n even, that is (1+ 1/n) which goes to 1.

    For n odd, that is -(1+1/n) which goes to -1.

    Those are the only two subsequential limits.
     
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