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I cant understand this explanation of limsup

  1. Jan 8, 2009 #1
    regarding this defintion

    http://img515.imageshack.us/img515/5666/47016823jz1.gif [Broken]

    i was told that
    Remember that if [itex]x_n[/itex] is bounded then [itex]\limsup x_n = \lim \left( \sup \{ x_k | k\geq n\} \right)[/itex].
    The sequence, [itex]\sup \{ x_k | k\geq n\}[/itex] is non-increasing, therefore its limits is its infimum.
    Thus, [itex]\limsup x_n = \inf \{ \sup\{ x_k | k\geq n\} | n\geq 0 \} [/itex][/quote]

    i cant understand the first part

    why he is saying that
    [itex]\sup \{ x_k | k\geq n\}[/itex]
    is not increasing.
    you are taking a bounded sequence and you get one number
    which is SUP (its least upper bound)
    thats it.
    no more members

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 8, 2009 #2


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    But [itex]\left{x_k|k\ge n}[/itex] is not a single sequence- it is a different sequence for every different n.

    For example, if [itex]{x_n= (-1)^n/n}= {-1, 1/2, -1/3, 1/4, -1/5, ...} then
    [itex]sup{x_k|k\ge 1}[/itex] is the largest of {-1, 1/2, -1/3, 1/4, -1/5, ...} which is 1/2. [itex]sup{x_k|k\ge 2}[/itex] is the largest of {1/2, -1/3, 1/4, -1/5, ...}, again 1/2. [itex]sup{x_k|k\ge 3}[/itex] is the largest of {-1/3, 1/4, -1/5, ...}, which is 1/4. Similarly, [itex]sup{x_k|k\ge 4}[/itex] is also 1/4 but [itex]sup{x_k|k\ge 5}[/itex] is 1/6, etc.
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