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Math Help - sequence series

  1. #1
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    sequence series

    sum of

    1/2+1/4+1/6+1/8... n terms
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  2. #2
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    1/2 + 1/4 + 1/8 + ...
    = (1/2)/(1 - 1/2)
    = 1

    This is sum to infinity of a geometric progression, with a = 1/2, r = 1/2
    S(infinity) = a/(1 - r)
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  3. #3
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     \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots = \sum_{i=1}^{n} \frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2}\times \frac{1}{i}= \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i}

    this follows http://en.wikipedia.org/wiki/Harmoni...s_(mathematics)
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  4. #4
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    Quote Originally Posted by gohpihan View Post
    1/2 + 1/4 + 1/8 + ...
    = (1/2)/(1 - 1/2)
    = 1

    This is sum to infinity of a geometric progression, with a = 1/2, r = 1/2
    S(infinity) = a/(1 - r)
    this maybe true but is not what the OP asked.
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  5. #5
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    Quote Originally Posted by pickslides View Post
     \displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots = \sum_{i=1}^{n} \frac{1}{2i} = \sum_{i=1}^{n}\frac{1}{2}\times \frac{1}{i}= \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i}

    this follows http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

    i have got the solution graphically as 1.2 but how to do it with zeta function
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  6. #6
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    As pickslides said, that is the same as \frac{1}{2}\sum_{i=1}^n \frac{1}{i}, the "harmonic series".
    Since \zeta(n) is defined as \sum_{i=1}^\infty \frac{1}{i^n}, \sum_{i=1}^\infty \frac{1}{n}= \zeta(1) and so \sum_{i=1}^\infty \frac{1}{2i}= \frac{1}{2}\zeta(a) but I do not know of any way to write the partial sums.
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