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Math Help - Proof for convergence of absolute value.

  1. #1
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    Proof for convergence of absolute value.

    Prove: |a_n| \rightarrow 0 \Rightarrow a_n \leftarrow 0

    So what I have is:

    0<||a_n|-0|< \epsilon

    0<|a_n|< \epsilon

    now a_n \leq |a_n| \therefore 0<a_n \leq |a_n|< \epsilon \therefore 0<a_n < \epsilon

    I'm not quit sure that this is right
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by lllll View Post
    Prove: |a_n| \rightarrow 0 \Rightarrow a_n \rightarrow 0

    So what I have is:

    0<||a_n|-0|< \epsilon

    0<|a_n|< \epsilon

    now a_n \leq |a_n| \therefore 0<a_n \leq |a_n|< \epsilon \therefore 0<a_n < \epsilon

    I'm not quit sure that this is right
    Hmm you have to separate the cases when a_n<0 and a_n>0

    More simply :
    |a_n| \to 0 \Leftrightarrow \forall \epsilon >0, ~ \exists N \in \mathbb{N}, ~ \forall n>N ~:
    ||a_n|-0|< \epsilon

    ||a_n||< \epsilon

    |a_n|< \epsilon

    |a_n-0|< \epsilon

    and that's the exact definition of a_n \rightarrow 0
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  3. #3
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    The sequence of formulas is as follows:

    Lim |a_n|=0 <====> \forall\epsilon[ε>0------> \exists k( kεN & \forall n( n \geqk--------> ||a_n|-0|< \epsilon))] <=====>


    \forall\epsilon[ε>0------> \exists k( kεN & \forall n( n \geqk--------> |a_n-0|< \epsilon))] <======> Lim (a_n)=0


    As n-----> OO ( infinity).

    The above equivalences hold because:


    ||a_n|-0|< \epsilon <=====> |a_n-0|< \epsilon, since:


    ||a_n|-0| = |a_n-0|,
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