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Math Help - Sequence Convergence

  1. #1
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    Sequence Convergence

    Recall that a sequence ( a_n) converges to L if and only if

     <br />
\forall \epsilon >0 \exists N \in \mathbb{N} \forall n \geq N  |a_n - L|< \epsilon<br />

    Prove that if a_n \rightarrow L then |a_n| \rightarrow |L|.

    Attempt:
    Suppose a_n \rightarrow L, then \lim_{x-> \infty} a_n = L

    Then we know -a_n \leq a_n \leq a_n

    The limit of the two outside terms is L, and hence the limit of a_n is L by the squeezing theorem for sequences.

    Conversely we can say: suppose |a_n| \rightarrow L then

    -|a_n| \leq a_n \leq |a_n|

    etc...

    But I doubt this is right or even the right approach. Can anyone show me a better and more rigorous proof?
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  2. #2
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    Quote Originally Posted by Roam View Post
    Prove that if a_n \rightarrow L then |a_n| \rightarrow |L|.
    It is really simple. This is all you need.
    \left| {\left| {a_n } \right| - \left| L \right|} \right| \leqslant \left| {a_n  - L} \right|
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  3. #3
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    That was really simple! Much simpler than what I was thinking:

    1) Suppose L is positive. Then you can take \epsilon = L/2 to show that for large enough n, a_n> 0 so |a_n|= a_n and the sequence |a_n| converges to L.

    2) Suppose that L is negative. Then you can take \epsilon= -L/2 to show that for large enough n, a_n< 0 so |a_n= -a_n and the sequence |a_n| converges to -L.

    3) Suppose that L= 0. Then [tex]||a_n|- |L||= ||a_n||= |a_n| = |a_n- L|.
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