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If a sequence is eventually bounded then it is bounded

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi, I've been solving Calculus Deconstructed by Nitecki and I've been confused by a particular lemma in the book. Namely:

    If a sequence is eventually bounded, then it is bounded:
    that is, to show that a sequence is bounded, we need only find a number
    γ ∈ R such that the inequality

    [tex]|x_i| < \gamma \ \text{holds for all i ≥ K, for some K}[/tex]


    2. Relevant equations
    3. The attempt at a solution

    However if we consider the sequence ##\tan {\frac{\pi}{n}} ##, with n starting at 1. For terms when n is very high, we can find some value for ##\gamma## for which the condition holds and thus the sequence is bounded (according to the lemma). Yet at n=1, ##\tan {\frac{\pi}{n}} = \infty##, so the sequence cannot be bounded.

    My questions are:

    1. Is my reasoning at all correct in the first place ?

    2. It is easy to see that my sequence does have a lower bound but no upper bound, so does the lemma only refer to one bound (But doesn't bounded mean bounded in both directions?)

    Thanks for reading!!
     
  2. jcsd
  3. Jun 16, 2015 #2

    Ray Vickson

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    Homework Helper

    You don't have a valid sequence, so that is not a counterexample to anything.

    Anyway, at ##n = 1## we have ##\tan(\pi) = 0##, so that is OK. The trouble occurs at ##n = 2##.

    But, I say it again: you don't have a valid sequence of real numbers with this example.
     
  4. Jun 16, 2015 #3
    .Oh I see, thats very illuminating thank you. Of course at n=2, the the element of the sequence is undefined and it is not infinite at all. I feel silly for making such a basic mistake now.
     
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