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Math Help - geometric series

  1. #1
    Junior Member
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    geometric series

    Problem 1

    n=0 to infinity (-1/2)^n (x-3)^n

    How would you find the value of x for which the given geometric series converges and how would you find the sum of the series(as a function of x) for the value of x?

    Problem 2

    How would you find the values of b for which

    1+ e^b+ e^2b+ e^3b+.....=9
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  2. #2
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    GP

    Hello TAG16
    Quote Originally Posted by TAG16 View Post
    Problem 1

    n=0 to infinity (-1/2)^n (x-3)^n

    How would you find the value of x for which the given geometric series converges and how would you find the sum of the series(as a function of x) for the value of x?
    (-\tfrac{1}{2})^n(x-3)^n = (-\tfrac{1}{2}(x-3))^n =\left(\frac{3-x}{2}\right)^n

    So the series converges if -1 < \frac{3-x}{2}<1

    \Rightarrow -2<3-x<2

    \Rightarrow -5<-x<-1

    \Rightarrow 5>x>1

    The sum to infinity S= \frac{a}{1-r}, where a = 1 and r = \frac{3-x}{2}

    \Rightarrow S = ... you can finish it now.

    Problem 2

    How would you find the values of b for which

    1+ e^b+ e^2b+ e^3b+.....=9
    This is a GP where a = 1 and r = e^b. So the sum to infinity S= \frac{a}{1-r} = \frac{1}{1 - e^b} = 9

    Solve this equation for b. (The answer will involve \text{ln}.)

    Grandad
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