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Real analysis: Limit superior proof

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Let (an) be a sequence of real numbers. Then we define
    lim [sup{an: n≥k}] = lim sup an

    (note: sup{an: n≥k} = sup{ak,ak+1,ak+2,...} = bk
    (bk) is itself a sequence of real numbers, indexed by k)

    Let a=lim sup an.
    Then for all ε>0, there exists N such that if n≥N, then an<a+ε

    2. Relevant equations

    3. The attempt at a solution
    I am trying to prove the theorem.
    Let bk=sup{an: n≥k}.
    By the definition of "a" as a limit of supremums (bk->a), we have that
    for all ε>0, there exists an integer N such that if k≥N, then
    |sup{an: n≥k} - a| = |bk -a|< ε
    => sup{an: n≥k} < a + ε
    By definition of an upper bound, sup{an: n≥k} ≥ an if n≥k.
    So the above shows that if k≥N and n≥k, then an< a+ ε.

    Now I am stuck...how should I continue? (I need to prove that: for all ε>0, there exists N such that if n≥N, then an<a+ε. But from my work so far, I'm really feeling hopeless)

    There are so many different subscripts (e.g. n and k) to keep track of that I am really puzzled and frustrated now.

    Can someone please help me out and show me the correct way to prove the theorem?
    Any help is very much appreciated!
    Last edited: Jan 16, 2010
  2. jcsd
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