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Math Help - Simple Geometric Series

  1. #1
    Member WhoCares357's Avatar
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    Simple Geometric Series

    \sum_{k=1}^\infty (\frac{2}{3})^{k+2}

    This is what I did.
    \sum_{k=1}^\infty \frac{4}{9}*(\frac{2}{3})^{k}
    a=\frac{4}{9}, r=\frac{2}{3}<1
    Threfore series converges to
    \frac{4}{3}

    This seems correct to me. However when I try to solve it thorugh maple (software) it outputs the result of \frac89. Did I do something wrong?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by WhoCares357 View Post
    \sum_{k=1}^\infty (\frac{2}{3})^{k+2}

    This is what I did.
    \sum_{k=1}^\infty \frac{4}{9}*(\frac{2}{3})^{k}
    a=\frac{4}{9}, r=\frac{2}{3}<1
    Threfore series converges to
    \frac{4}{3}

    This seems correct to me. However when I try to solve it thorugh maple (software) it outputs the result of \frac89. Did I do something wrong?
    Your value of a is incorrect. As the sum starts at k=1 you'd need to do

    \frac{4}{9} \times \left(\frac{2}{3}\right)^1 = \frac{8}{27}

    Putting in this value of a should give \frac{8}{9}
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  3. #3
    Member WhoCares357's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Your value of a is incorrect. As the sum starts at k=1 you'd need to do

    \frac{4}{9} \times \left(\frac{2}{3}\right)^1 = \frac{8}{27}

    Putting in this value of a should give \frac{8}{9}
    Can you explain why? Wouldn't that force the series to start from 2?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by WhoCares357 View Post
    Can you explain why? Wouldn't that force the series to start from 2?
    As I understood it as it's the first term a would be the value of f(k) at k=1 as that's where the sum starts from.

    I don't think the series starts at 2 as a is the initial value before anything is added
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  5. #5
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    Just a question I am new to series, because this geometric series starts at n=1 would you not try to get the power in terms of n - 1?
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