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Math Help - Infinite Geometric Series

  1. #1
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    Infinite Geometric Series

    Find the sum of the given infinite geometric series if possible.
    The Answer is 0.5. How do I solve this to get to the answer


    \sum_{i=1}^{\infty} 2(\frac{1}{5})i


    Thanks
    Last edited by ToXic01; May 9th 2010 at 02:57 PM.
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  2. #2
    Junior Member slovakiamaths's Avatar
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    i can't understand your Que
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  3. #3
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    Question

    Quote Originally Posted by ToXic01 View Post
    A question why is the answer 0.5

    n=infinity
    SIGMA 2(1/5)i
    r=1

    Thanks
    ToXic01: I cannot understand either. Can you clarify?
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  4. #4
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by ToXic01 View Post
    A question why is the answer 0.5

    infinity
    SIGMA 2(1/5)i
    r=1

    Thanks
    \sum_{i=1}^{\infty} 2(\frac{1}{5})^i = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\} = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}

    aNon1
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  5. #5
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    Apparently I don't know how to use latex for symbols I had to copy Anonymous1 code to fix the problem. See my op post for the correct question. Thanks
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  6. #6
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    Quote Originally Posted by ToXic01 View Post
    Apparently I don't know how to use latex for symbols I had to copy Anonymous1 code to fix the problem. See my op post for the correct question. Thanks
    Do you understand how I arrived at the solution?
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  7. #7
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    Quote Originally Posted by Anonymous1 View Post
    Do you understand how I arrived at the solution?
    Because of my incorrect question, the ^i is not suppose to be like that. So I am not sure if you had the correct question for me to understand.However, to answer you question I am still not sure how you came to that conclusion

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  8. #8
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    The edited question is not a geometric series, and becomes infinitely large.

    \sum_{i=1}^{\infty} 2(\frac{1}{5})i = \frac{2}{5}\sum_{i=1}^{\infty}i \to \infty

    On the other hand...

    \sum_{i=1}^{\infty} 2(\frac{1}{5})^i

    ...is a geometric series that converges to \frac{1}{2}.
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  9. #9
    Super Member Anonymous1's Avatar
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    The geometric series formula is \sum_{i=0}^{\infty} a(r)^i = a\frac{1}{1-r}

    Your sum is:

    \sum_{i=1}^{\infty} 2(\frac{1}{5})^i =2\sum_{i=1}^{\infty} (\frac{1}{5})^i

    Notice the indices go from i=1 to \infty. We want it to go from i=0 to \infty so that we can use the geometric series formula.

    The i=0 term is (\frac{1}{5})^0 = 1.

    So I both added 1 (by changing the sum from [1,\infty] to [0,\infty] ) and subtracted 1.

    = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\}

    Now, apply the formula for the geometric series on [\sum_{i=0}^{\infty} (\frac{1}{5})^i] where r=\frac{1}{5} and a=1.

    = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}
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  10. #10
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    Why not learn to post in symbols? You can use LaTeX tags.
    [tex] \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}[/tex] gives  \sum\limits_{k = 1}^\infty  {\frac{2}{{5^k }}}  = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}
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  11. #11
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    yea i doubled checked
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  12. #12
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    can i ask how you got the ratio 1/5
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  13. #13
    Junior Member slovakiamaths's Avatar
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    \sum\limits_{k = 1}^\infty  {\frac{2}{{5^k }}}=\frac{2}{5^1}+\frac{2}{5^2}+\frac{2}{5^3}+....  ...................
    here a=2/5 and r=\frac{2}{5^2}\div\frac{2}{5^1}=\frac{1}{5}
    \sum\limits_{k = 1}^\infty  {\frac{2}{{5^k }}}  = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}
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