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Math Help - Infinite Series

  1. #1
    Member rtblue's Avatar
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    Infinite Series

    Find the sum of:

    1+\frac{1}{3}+\frac{2}{9}+\frac{1}{9}+\frac{4}{81}  +\frac{5}{243}....

    I have tried to group the terms, to create two geometric series, but that does not work. I can see no pattern from one term to the next, although i have noted:

    1+\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{  4}{3^4}+\frac{5}{3^5}....

    but I do not think this observation helps much. Any assistance is appreciated.
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  2. #2
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    I would write it as \displaystyle1 + \sum\limits_{k = 1}^\infty  {\frac{k}{{3^k }}} .

    Recall that if \displaystyle\sum\limits_{k = 1}^\infty  {kx^k }  = \frac{x}{{\left( {1 - x} \right)^2 }},\;\left| x \right| < 1.

    Let x=\frac{1}{3}
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  3. #3
    MHF Contributor chisigma's Avatar
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    You are right!... the series can be written as...

    \displaystyle S= 1 + \sum_{n=1}^{\infty} n\ (\frac{1}{3})^{n} (1)

    If we introduce the function...

    \displaystyle \sigma(x)= 1 + \sum_{n=1}^{\infty} n\ x^{n}= 1 + x\ \sum_{n=1}^{\infty} n\ x^{n-1} = 1+ x\ \frac{d}{dx} \sum_{n=0}^{\infty} x^{n}=

    \displaystyle 1 + x\ \frac{d}{dx} \frac{1}{1-x} = 1+\frac{x}{(1-x)^{2}} (2)

    ... is easy to derive that is...

    \displaystyle S=\sigma(\frac{1}{3}) = \frac{7}{4} (3)

    Kind regards

    \chi \sigma
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  4. #4
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    Hello, rtblue!

    Here is an elementary solution . . .


    Find the sum of: . \displaystyle S \;=\;1+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\fr  ac{4}{3^4}+\frac{5}{3^5}\ + \hdots

    \begin{array}{ccccccc}<br />
\text{We have:} & S &=& 1 + \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \hdots & [1] \\<br />
\text{Multiply by}\frac{1}{3}\!: & \frac{1}{3}S &=& \quad\;\;\;\frac{1}{3} + \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \hdots & [2]  \\\end{array}

    Subtract [1] - [2]: . \frac{2}{3}S \;=\;1 + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \hdots

    . . . . . . . . . . . . . \frac{2}{3}S \;=\;1 +\frac{1}{3^2}\underbrace{\left(1 + \tfrac{1}{3} + \tfrac{1}{3^2} + \tfrac{1}{3^3} + \hdots \right)}_{\text{Infinite series}}\;\;[3]


    The infinite series has the sum: . \dfrac{1}{1-\frac{1}{3}} \:=\:\dfrac{1}{\frac{2}{3}} \:=\:\frac{3}{2}


    Substitute into [3]: . \frac{2}{3}S \;=\;1 + \frac{1}{9}\left(\frac{3}{2}\right) \;=\;1 + \frac{1}{6} \;=\;\frac{7}{6}

    . . . . . . Therefore: . . S \;=\;\frac{7}{4}

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