1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lim sup and lim inf fog

  1. Oct 4, 2005 #1
    Hi,
    I'm having trouble understanding why a sequence converges if and only if lim sup=lim inf. I think about the sequence {1/n} this sequence converges to 0, but the lim sup is 1. How is the limsup 0? What am I missing?
    HELP
    CC
     
  2. jcsd
  3. Oct 4, 2005 #2

    TD

    User Avatar
    Homework Helper

    Could you please be a bit more clear? What do you mean exactly with "lim sup" and "lim inf". The 'limit' of the 'supremum' and 'infinity'?

    As you say, {1/n} converges to 0 (i.e. the sequence goes to 0 as n goes to infinity), but what do you mean with "limsup"?
     
  4. Oct 4, 2005 #3
    that's exactly my quandry.
    There's a theorem in the book that says
    "If {s} is a sequence of real numbers, and if limsup s = liminf s =L where L is in R, then {s} is convergent and lim s = L."

    So I don't understand how in the case of {1/n} this holds. 1/n is a convergent sequence and it's limit is 0, but looking at the theorem there, it seems that limsup s must be zero also. But limsup {1/n} is 1. What am I missing?

    HELP
     
  5. Oct 5, 2005 #4
    lim sup is the limit of the supremum for large N, right? lim sup sn = lim as N-> infinity of the set {sn : n > N}.

    So begin listing some terms of sn: 1, 1/2, 1/3, 1/4, 1/5, 1/6, ...

    If N = 1, then n must be at least 2. So we ignore the first term, and look at the sequence 1/2, 1/3, 1/4, 1/5, 1/6, ..... The sup in this case is 1/2.

    If N = 2, then n must be at least 3. So we ignore the first two terms, and look at the sequence 1/3, 1/4, 1/5, 1/6, 1/7, ... The sup in this case is 1/3.

    If N = 3, then n must be at least 4. So we ignore the first three terms, and look at the sequence 1/4, 1/5, 1/6, 1/7, 1/8. Now the sup is 1/4.

    We must look at the lim sup. So we need the LIMIT of the supremum. Of course the limit of the supremum is going to be 0.

    Similarly, the lim inf is 0:

    If N = 1, then n must be at least 2, so we look at the sequence 1/2, 1/3, 1/4, 1/5, ...
    and you can see that the lim inf is 0.

    Thus lim inf sn = lim sup sn = 0.

    I think that's the right way to look at it!
     
  6. Oct 5, 2005 #5
    OOOOOOOOOOOO
    Now I see what they mean. The fog has lifted!
    THANKYOU VERY VERY MUCH
     
  7. May 21, 2015 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    More correctly, "lim sup" of a sequence is the supremum (upper bound) of all subsequential limits. And a sequence converges if and only if all subsequences converge to the same limit.

    You are confusing "lim sup" of a sequence with "sup" of a set. The supremum of the set {1/n} is the largest member, 1, but since the sequence converges to 0, all subsequences also converge to 0. The set of all "subsequential limits" is {0} and both sup and inf of that set is 0.
     
  8. May 21, 2015 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Posters have been patiently waiting over 9 years for that assistance, HallsofIvy.

    I hope it's appreciated; if not by them, then at least by others now that you have revitalised the thread. :wink:
     
  9. May 21, 2015 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How in the world did I manage to get into "2005"? Time travel?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Lim sup and lim inf fog
  1. Lim inf and Lim sup (Replies: 9)

  2. Lim Sup, Lim Inf Proof (Replies: 4)

  3. Lim sup An lim inf An (Replies: 2)

Loading...