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Math Help - Is the sequence Convergent?

  1. #1
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    Is the sequence Convergent?

    Hi, my answer is True. Am i correct?

    A sequence {an} satisfies an+1= 2an/(1+an) for n=1,2,....
    If an=2, the sequence is monotonic decreasing and bounded, so it is convergent.

    (a) True
    (b) False

    Thanks for checking (:
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  2. #2
    Newbie Esteban's Avatar
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    Re: Is the sequence Convergent?

    Quote Originally Posted by weijing85 View Post
    Hi, my answer is True. Am i correct?

    A sequence {an} satisfies an+1= 2an/(1+an) for n=1,2,....
    If an=2, the sequence is monotonic decreasing and bounded, so it is convergent.

    (a) True
    (b) False

    Thanks for checking (:
    that's what exercise says? because, if a_n=2 it's a constant sequence, then it's trivially convergent
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  3. #3
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    Re: Is the sequence Convergent?

    Quote Originally Posted by weijing85 View Post
    A sequence {an} satisfies an+1= 2an/(1+an) for n=1,2,.... If an=2, the sequence is monotonic decreasing and bounded, so it is convergent.
    (a) True
    (b) False
    I suspect there is a typo in the post,
    Is it not a_1=2~? If so, use induction to show a_n>a_{n-1}>1.

    Hint: note that if a>1 then a<a^2.

    To find the limit solve L=\frac{2L}{1+L}.
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  4. #4
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    Re: Is the sequence Convergent?

    Thanks for the correction. Is a1 =2.
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  5. #5
    Junior Member Nehushtan's Avatar
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    Re: Is the sequence convergent?

    Alternatively you can prove by induction that a_n=\frac{2^n}{2^n-1} and take it from there. (The formula is easily guessed after trying a few values of n.)
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  6. #6
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    Re: Is the sequence Convergent?

    So the answer is true?
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  7. #7
    Junior Member Nehushtan's Avatar
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    Re: Is the sequence convergent?

    Well, thatís for you to find out, isnít it? I mean, it may be that you find the formula a_n=\frac{2^n}{2^n-1} easier to work with; if not, you can always stick to the original recursive formula.
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