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Math Help - Geometric Series

  1. #1
    Senior Member slevvio's Avatar
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    Talking Geometric Series

    I was having trouble with this question and wondered if I could get some help with it ?

    How many terms of the geometric series  1 + \frac{2}{3} + \frac{4}{9} + ... must be taken so that the sum differs from 3 by less than  10^{-3} ?

    I started with this inequality:

    <br />
3 - 10^{-3} < \frac{(\frac{2}{3})^n - 1}{\frac{2}{3} - 1} < 3 + 10^{-3}<br />

    and ended up with this:

    <br />
\frac{ln(\frac{-1}{3000})}{ln(\frac{2}{3})} < n < \frac{ln(\frac{1}{3000})}{ln(\frac{2}{3})} = 19.74<br />

    Of course you cannot take the log of a negative number so how do I work out what n must be greater than? Also sorry I have no idea how to do that maths code thing so it might look a bit messy, Thanks
    Last edited by slevvio; November 4th 2007 at 10:13 AM.
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  2. #2
    Member OnMyWayToBeAMathProffesor's Avatar
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    codes

    here is the page the explains codes.

    3 - 10^(-3 < (2/3)^n - 1 )/(2/3 - 1) < 3 + 10 ^-3 would be

    3 - 10^ < \frac{(\frac{2}{3})^n - 1 )}{(\frac{2}{3} - 1) < 3 + 10^  (-3}

    u have to play with it.
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  3. #3
    Super Member

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    Hello, slevvio!

    I discovered that there is a strange step in this solution . . .


    How many terms of the geometric series  1 + \frac{2}{3} + \frac{4}{9} + \cdots
    must be taken so that the sum differs from 3 by less than  10^{-3} ?

    The sum of the first n terms is: . S_n \;=\;\frac{1 - (\frac{2}{3})^n}{1-\frac{2}{3}} \;=\;3\left[1-\left(\frac{2}{3}\right)^n\right]

    We want: . \left|\,3\left[1-\left(\frac{2}{3}\right)^n\right] - 3\,\right| \;<\;0.001

    Then: . -0.001 \;<\;3\left[1 - \left(\frac{2}{3}\right)^n\right] - 3 \;<\;0.001

    Divide by 3: . -\frac{0.001}{3} \;<\;1 - \left(\frac{2}{3}\right)^n - 1 \;<\;\frac{0.001}{3}\quad\Rightarrow\quad -\frac{0.001}{3} \;<\:\underbrace{-\left(\frac{2}{3}\right)^n \;<\;\frac{0.001}{3}}

    The right half of the inequality can be discarded.
    . . (Of course, a negative fraction is less than a positive fraction.)

    So we have: . -\frac{0.001}{3} \;<\;-\left(\frac{2}{3}\right)^n \quad\Rightarrow\quad<br />
-\left(\frac{2}{3}\right)^n \;>\;-\frac{0.001}{3}

    Multiply by -1: . \left(\frac{2}{3}\right)^n \;< \;\frac{0.001}{3}

    Take logs: . \ln\left(\frac{2}{3}\right)^n \;<\;\ln\left(\frac{0.001}{3}\right)\quad\Rightarr  ow\quad n\cdot\ln\left(\frac{2}{3}\right) \;<\;\ln\left(\frac{0.001}{3}\right)


    Divide both sides by \ln\left(\frac{2}{3}\right) . . . note that this quantity is negative.

    . . n \;> \;\frac{\ln\left(\frac{0.001}{3}\right)}{\ln\left(  \frac{2}{3}\right)} \;= \;19.7461...


    Therefore: .  n \:\geq\:20


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  4. #4
    Senior Member slevvio's Avatar
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    Thank you you certainly are a super member!
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