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Math Help - Sequence Series Question

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    Sequence Series Question

    Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)
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    Quote Originally Posted by acevipa View Post
    Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)
    Solve \frac{1}{1-w} = \frac{a}{1 - \frac{1}{w}} for w subject to the restriction \left| \frac{1}{w}\right| < 1 (a is the first term in the series).
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    Quote Originally Posted by acevipa View Post
    Find a geometric series with common ration 1/w that has limiting sum 1/(1-w)
    hey mate,

    good question

    As Im sure you well aware a geometric series a(n) is defined so that
    a(n+1) = r * a(n), where r is the common ratio
    here you have given that r = 1/w
    thus,
    a(n+1) = (1/w)a(n)
    assuming a(0) = a you then have a(n) = (1/w)^n * a(0) = a * (1/w)^n

    furthermore you have stated the the limiting sum convergences to 1/(1-w), i.e.

    sum(a(n), n = 0..inf) = 1/(1-w)
    or
    sum(a * (1/w)^n, n = 0..inf) = 1/(1-w)
    since a is a constant, bring it out the front of the sum
    thus
    = a * sum ( (1/w)^n, n = 0...inf) = 1/(1-w)

    The sum component is known as the sum of a geometric series, whereby

    a(n) is geometric, then
    sum(a(n), n = 0...inf)) = 1/(1-r)
    where r is the common ratio, here r = 1/w

    Thus the 'sum' component of the previous equation can be replaced with
    1/ (1- (1/w)) = w/(w-1)

    hence we obtain

    a * w/(w-1) = 1/(w-1)
    hence a = 1/w

    Thus, a(n) = a * (1/w)^n = (1/w)^(n+1), n = 0... inf


    Hope this helps,

    Let me know if you require any further clarrification and if I have made any mistakes,

    Regards,

    David
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