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Math Help - Sum of Series

  1. #1
    Senior Member polymerase's Avatar
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    Sum of Series

    Find the sum of the series \sum _{n=3}^{\infty} \frac{n}{3^n}

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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by polymerase View Post
    Find the sum of the series \sum _{n=3}^{\infty} \frac{n}{3^n}

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    I will explain later but the answer is \frac{7}{36}
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  3. #3
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    Quote Originally Posted by polymerase View Post
    Find the sum of the series \sum _{n=3}^{\infty} \frac{n}{3^n}

    Thanks
    For |x|<1:

    \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \implies \left( \sum_{n=0}^{\infty} x^n \right)' = \frac{1}{(1-x)^2} \implies \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}

    Use that to solve the problem.
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  4. #4
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    Hello, polymerase!

    Find the sum of the series: . \sum _{n=3}^{\infty} \frac{n}{3^n}

    We have: . . . . . S \;=\;\frac{3}{3^3} + \frac{4}{3^4} + \frac{5}{3^5} + \frac{6}{3^6} + \cdots

    Multiply by \frac{1}{3}\!:\;\;\frac{1}{3}S \;=\qquad\quad\frac{3}{3^4} + \frac{4}{3^5} + \frac{5}{3^6} + \hdots

    Subtract:. . . . . \frac{2}{3}S \;=\;\frac{3}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \frac{1}{3^6} + \cdots


    We have: . \frac{2}{3}S \;=\;\frac{1}{9} + \frac{1}{3^4}\underbrace{\left(1 + \frac{1}{3} + \frac{1}{3^2} + \hdots\right)}_{\text{geometric series}}

    . . The geometric series has a sum of: . \frac{1}{1-\frac{1}{3}} \:=\:\frac{3}{2}

    So we have: . \frac{2}{3}S \;=\;\frac{1}{9}+ \frac{1}{81}\!\left(\frac{3}{2}\right) \;=\;\frac{7}{54}


    Therefore: . S \;=\;\frac{3}{2}\!\left(\frac{7}{54}\right) \;=\;\frac{7}{36}

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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Just wondering

    Quote Originally Posted by ThePerfectHacker View Post
    For |x|<1:

    \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \implies \left( \sum_{n=0}^{\infty} x^n \right)' = \frac{1}{(1-x)^2} \implies \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}

    Use that to solve the problem.
    but for this case then would it be \sum_{n=3}^{\infty}nx^{n}=\frac{x^3}{(1-x)^2}
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