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Thread: 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 $\displaystyle \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 $\displaystyle \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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    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 $\displaystyle 3^{-k}=\left( \frac{1}{3} \right)^k$.

    For your problem, you want to find a two constant $\displaystyle a,r$ such that : $\displaystyle \frac{a}{1-r}=3$
    choose any value for r such that $\displaystyle |r|<1$ (to make the series coverges) then put this value in the last equation and solve it for a.
    Once you have $\displaystyle a,r$ 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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