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Math Help - Question about bounded sequences and limits

  1. #1
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    Question about bounded sequences and limits

    Hi everyone,

    I've got a problem on a practice test that I can't quite get my head around.

    Let the sequence a subscript n, going from 1 to infinity, be bounded above by M. If the sequence goes to L as n goes to infinity, show (using an epsilon - N argument) that L <= M.

    I'm given the hint: Assume to the contrary that L > M, then (L-M)>0 so you may take epsilon = (L-M)/2 in your epsilon - N argument.

    I've gotten to the stage where I have (L+M)/2 < a subscript n < (3L-M)/2 and am not sure how to proceed.
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  2. #2
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    Quote Originally Posted by pineapple89 View Post
    Hi everyone,

    I've got a problem on a practice test that I can't quite get my head around.

    Let the sequence a subscript n, going from 1 to infinity, be bounded above by M. If the sequence goes to L as n goes to infinity, show (using an epsilon - N argument) that L <= M.

    I'm given the hint: Assume to the contrary that L > M, then (L-M)>0 so you may take epsilon = (L-M)/2 in your epsilon - N argument.

    I've gotten to the stage where I have (L+M)/2 < a subscript n < (3L-M)/2 and am not sure how to proceed.

    So you took \epsilon=\frac{L-M}{2} ,and thus there exists N_\epsilon\in\mathbb{N}\,\,\,s.t.\,\,\,n>N_\epsilo  n\Longrightarrow |a_n-L|<\epsilon \Longleftrightarrow -\epsilon<a_n-L<\epsilon\Longrightarrow -\epsilon+L<a_n<\epsilon+L . But -\epsilon+L=\frac{L+M}{2}>\frac{M+M}{2}=M \Longrightarrow a_n>-\epsilon+L>M , and this contradicts the definition of M...

    Tonio
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