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Math Help - unsure of a proof

  1. #1
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    unsure of a proof

    Ok so the question is:
    Prove carefully that a convergent sequence is bounded.
    Here is my proof thus far, but i'm not really sure if it is correct\finished:
    |a_n| = |a_n - L + L| \leq |a_n - L| + |L|

    a_n is convergent, say to a limit L so:
    Let \epsilon > 0
    \exists N \in \mathbb{N} such that \forall n \geq N |a_n - L| < \epsilon.

    |a_n| \leq |a_n - L| + |L| < \epsilon + |L| \forall n \geq N.

    Is this enough to show that it is bounded?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes, but I will wrote a better proof...

    \{a_n\}^{\infty}_{n=1} convergent say to L hence by definition:

    Exist N\in\mathbb{N} so for all n>N: |a_n-L|<1.

    Let us mark M=max\{|a_1|,|a_2|,...,|a_N|,|L|+1\} and then if n\leq N then we have: |a_n|\lwq M

    And if n>N then:

    |a_n|\leq |L|+|a_n-L|<|L|+1\leq M

    Hence M is upper bound to sequence \{|a_n|\}^{\infty}_{n=1}, the conclusion is that \{a_n\}^{\infty}_{n=1} bounded.


    This is more elegant proof.
    Last edited by Also sprach Zarathustra; January 3rd 2011 at 02:19 PM.
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  3. #3
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    Here is the standard proof.
    If for every \epsilon>0 it is true that |a|<|b|+\epsilon then |a|\le|b|.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    I may be getting into semantics here, but wouldn't this sequence be a counterexample?

     a_n = \frac1{(n-1)^2} since  a_1 isn't defined?
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    I may be getting into semantics here, but wouldn't this sequence be a counterexample?
     a_n = \frac1{(n-1)^2} since  a_1 isn't defined?
    Most competent mathematicians would not be that picky.
    Now granted that is outside the standard class in real analysis, or some such course.
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