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Math Help - prove that if sequence (an) converges, use def of limit of sequence to prove...

  1. #1
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    prove that if sequence (an) converges, use def of limit of sequence to prove...

    that if an>=1 for all natural number n, then L >= 1.

    My answer :

    From the def of limit of sequence,

    L-a(n) < epsilon and a(n) - L < epsilon

    we know that a(n) >=1,

    L - a(n) < epsilon => L must be larger than or same with a(n) to keep epsilon positive.

    a(n) - L < epsilon => L must be larger than or same with a(n) to keep the equation on the left smaller than epsilon.

    Is this argument acceptable?
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  2. #2
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    Quote Originally Posted by serenalee View Post
    that if an>=1 for all natural number n, then L >= 1.

    My answer :

    From the def of limit of sequence,

    L-a(n) < epsilon and a(n) - L < epsilon

    we know that a(n) >=1,

    L - a(n) < epsilon => L must be larger than or same with a(n) to keep epsilon positive.

    a(n) - L < epsilon => L must be larger than or same with a(n) to keep the equation on the left smaller than epsilon.

    Is this argument acceptable?
    This is incorrect.

    In your proof you say that because L - a_n < \epsilon , L \geq a_n for \epsilon to be positive, however if L < a_n then L - a_n < 0 \Rightarrow L - a_n < \epsilon for all  \epsilon > 0

    My idea of proving this would be to assume, in contradiction, that L < 1 and then you can find some \epsilon that has no N such that for all n>N |a_n-L|<\epsilon. Can you follow from here?

    Spoiler:
    Proof:
    Assume by contradiction that L<1 and let \epsilon_0 = 1-L . Assume that there exists n_0 \in \mathbb{N} such that for all n>n_0 , |a_n - L| < \epsilon_0
    Then, |a_n - L| < 1-L \Rightarrow a_n - L < 1 - L \Rightarrow a_n<1 - contradiction.
    Last edited by Defunkt; August 28th 2009 at 04:24 AM.
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  3. #3
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    Hi
    the definition of limit says that for each \varepsilon > 0 there exists n_0 \in \mathbb{N} such that for each n > n_0 we have \left|L-a_n\right| < \varepsilon.
    Let's argue by contradiction: assume that L < 1. Set \varepsilon = \frac{1-L}{2}. Then, according to the definition, there must exist n_0 \in \mathbb{N} such that for each n > n_0 we have \left|L-a_n\right| < \varepsilon = \frac{1-L}{2}. But this means a_n < 1, a contradiction.
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