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Lim inf lim sup ratio question

  1. Dec 12, 2008 #1
    i am given a siquence An of positive numbers

    how to prove this expression:

    lim sup(1/An)= 1/(lim inf An)

    lim inf is the infimum of the group of the limits of the sequences
    lim sup is the supremum of the group of the limits of the sequences

    those groups of limits doesnt have to be connected to one another
    then so is their limits

    its like saying that inf Xn=1/sup Xn for every sequence

    ??
     
  2. jcsd
  3. Dec 12, 2008 #2

    HallsofIvy

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    Let A= lim inf An. Then given any [itex]\epsilon> 0[/itex] there exist a subsequence of An whose limit is within [itex]\epsilon[/itex] of A. What can you say about 1/An for that subsequence?

    What? Unless I am misunderstanding you, those "groups of limits" are exactly the same! They are both "the limits of all convergent subsequences".

    Do you mean every subsequence of a specific Xn? In any case, no this does not say anything like that.
     
  4. Dec 12, 2008 #3
    if i am given e>0 for which |An-A|<e
    so after some N values all the An are will be between A+e and A-e

    so all the limits of every sub sequence of An will be between A+e and A-e
    so when i write 1/(An) :
    after N values the values of the limits will go from 1/(A+e) to 1/(A-e)

    so 1/lim inf(An) will return 1/inf( |An-A| group)
     
    Last edited: Dec 12, 2008
  5. Dec 12, 2008 #4
    the lim sup(1/An) is
    the supremum of the group (1/|An-A| )

    the inf is the lowest denominator so it will live the highest number possible

    and the sup is by definition will that the highest number

    so its a correct math proof??

    what mathematical expressions do i miss ?

    what bothers me is :
    my proof will work for a group of limits
    but we are dealing with An which is not
     
    Last edited: Dec 12, 2008
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