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Math Help - Another question about a limit of a sequence

  1. #1
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    Another question about a limit of a sequence

    Hey people.
    I need to prove or to contradict this claim :
    "If a sequence (a_n) converges to a finite limit and a_n != 0 for each n , then the limit of |a_(n+1)|/|a_n| as n->inf is smaller or equal to 1 , i.e
    lim(|a_(n+1)|/|a_n|) <=1

    I think that this claim is true . if a_n converges to a limit L, then of course a_(n+1) converges to L as well. Therefore, if L is not 0 , then lim(|a_(n+1)|/|a_n|) = |L|/|L| = 1

    The problem here is what if L is to prove this claim is L is 0.
    I think it should be done through the definition of a limit of a sequence.
    I tried to play with that a little , but I didn't manage to prove that.

    Any clues ?
    Thanks in advance.!
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  2. #2
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    ?? Try a_{n} = \frac{n}{1-n} for n > 1

    Not everything approaches from the top.
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  3. #3
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    for a_n = n/(1-n),
    a_(n+1) = (n+1)/(-n)
    therefore
    a_(n+1) / a_n = ((n+1)/(-n))/(n/(1-n)) = (-((1-n) (n+1))/n^2)

    And the limit of |(-((1-n) (n+1))/n^2)| as n->inf is 1
    Therefore this example does not contradict the claim.

    I think that the claim is true.
    I tried to prove that through the definition of limit of a sequence , but I didn't manage to.
    Last edited by Gok2; February 24th 2010 at 04:48 AM.
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  4. #4
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    Fair enough. Now prove it. Contradiction might be a useful approach.
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  5. #5
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    I tried to use contradiction, but the point is that I need to prove here two things : First , I need to prove that lim( |a_(n+1) / a_n| ) exists and finite, and only then I need to prove that this limit is smaller or equal to 1..
    I don't see how I can prove using contradiction that this limit exists and finite...

    Any clues, anyone?
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