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Math Help - how to prove that this series bounded and monotonic..

  1. #1
    MHF Contributor
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    how to prove that this series bounded and monotonic..

    Xn=(1-1/2)(1-1/4)..(1-(1/(2^n))

    i tried to prove that its monotonic
    by :
    1-1/(2^n) = (2^n-1)/2^n

    2^n -1 <2^n
    obviously its correct
    the numerator of each object is smaller then the denominator.

    what now??

    and how to prove that its bounded?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    Xn=(1-1/2)(1-1/4)..(1-(1/(2^n))

    i tried to prove that its monotonic
    by :
    1-1/(2^n) = (2^n-1)/2^n

    2^n -1 <2^n
    obviously its correct
    the numerator of each object is smaller then the denominator.

    what now??

    and how to prove that its bounded?
    Is there a reason you keep calling these series instead of products?

    You want to prove monotonicity and boundedness of \prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]. Boundedness is simple

    \forall{n}\in(1,\infty)~1-\frac{1}{2^k}<1\implies\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}1=1

    As for monotonicity you are making a mistake we are talking about the monotonicty of a_n, that is a_n=\prod_{k=1}^{n}b_k NOT \prod_{k=1}^{n}a_k. So we need to show that \prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]. Since both sides are positive for all n we have that

    \prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]\implies\displaystyle{\frac{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}<\frac{\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}}. Simplifying gives 1<\frac{1}{1-\frac{1}{2^{n+1}}} which is obviously true for all positive n.
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