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Thread: Proof for convergence of absolute value.

  1. #1
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    Proof for convergence of absolute value.

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

    So what I have is:

    $\displaystyle 0<||a_n|-0|< \epsilon $

    $\displaystyle 0<|a_n|< \epsilon $

    now $\displaystyle 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: $\displaystyle |a_n| \rightarrow 0 \Rightarrow a_n \rightarrow 0 $

    So what I have is:

    $\displaystyle 0<||a_n|-0|< \epsilon $

    $\displaystyle 0<|a_n|< \epsilon $

    now $\displaystyle 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 $\displaystyle a_n<0$ and $\displaystyle a_n>0$

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

    $\displaystyle ||a_n||< \epsilon$

    $\displaystyle |a_n|< \epsilon$

    $\displaystyle |a_n-0|< \epsilon$

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

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


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


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

    The above equivalences hold because:


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


    $\displaystyle ||a_n|-0|$ =$\displaystyle |a_n-0|$,
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