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Thread: 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


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


    Thanks
    Last edited by ToXic01; May 9th 2010 at 01:57 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
    $\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})^i = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\} = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}$

    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.

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

    On the other hand...

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

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

    Your sum is:

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

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

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

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

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

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

    $\displaystyle = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}$
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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 $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$
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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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    $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}}=\frac{2}{5^1}+\frac{2}{5^2}+\frac{2}{5^3}+.... ...................$
    here a=2/5 and $\displaystyle r=\frac{2}{5^2}\div\frac{2}{5^1}=\frac{1}{5}$
    $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$
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