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Math Help - a convergent geometric series...

  1. #1
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    a convergent geometric series...

    i need help finding a convergent geometric series that converges to 3.

    i know the formula for a geometric sreries is the sum cr^n, and that if the absolute value of r < 1 it converges to zero...but i dont know how to make one converge to 3.

    would this converge to 3? sum 3(2/3)^n, since 2/3 will go to zero and then the c value of 3 will make it converge to 3?

    thanks for any help
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  2. #2
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    Quote Originally Posted by andrewt328 View Post
    i need help finding a convergent geometric series that converges to 3.
    Does \sum\limits_{k = 0}^\infty  {2 \cdot 3^{ - k} } work?
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    Quote Originally Posted by andrewt328 View Post
    i need help finding a convergent geometric series that converges to 3.

    i know the formula for a geometric sreries is the sum cr^n, and that if the absolute value of r < 1 it converges to zero...but i dont know how to make one converge to 3.

    would this converge to 3? sum 3(2/3)^n, since 2/3 will go to zero and then the c value of 3 will make it converge to 3?

    thanks for any help
    your gp is
    3[(2/3)+(2/3)^2+.......]=2+2(2/3)+2(2/3)^2+2(2/3)^3+....

    sum to infinity is a/1-r so we have 2/(1-2/3)=6

    can you see how to amewnd your gp to get a sum of 3?
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    Quote Originally Posted by Plato View Post
    Does \sum\limits_{k = 0}^\infty {2 \cdot 3^{ - k} } work?
    i dont think that will work...but thats not a geometrical series since it is to the -k...right?


    and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know
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  5. #5
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    Quote Originally Posted by andrewt328 View Post
    i dont think that will work...but thats not a geometrical series since it is to the -k...right?


    and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know
    But 3^{-k}=\left( \frac{1}{3} \right)^k.

    For your problem, you want to find a two constant a,r such that : \frac{a}{1-r}=3
    choose any value for r such that |r|<1 (to make the series coverges) then put this value in the last equation and solve it for a.
    Once you have a,r you can put them in the general form of the geometric series to get the desired series.
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    Quote Originally Posted by andrewt328 View Post

    and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know

    i wrote out the series, it is
    2+2(2/3)+2(2/3)^2+2(2/3)^3+....

    from there we can read off the first term (my a,presumably your c) as 2 and common ratio 2/3


    gp mean geometric progression which is what your question is about.
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  7. #7
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    thanks for the help guys...
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