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Math Help - Integral test for series problem

  1. #1
    Senior Member mollymcf2009's Avatar
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    Integral test for series problem

    I am not sure how I need to solve this. Can someone get me started?

    How many terms of the following series would you need to add to find its sum to within 0.01?

    \sum^{\infty}_{n=2} \frac{4}{n(ln(n))^9}

    a) n > e^{(25)^{\frac{1}{4}}}

    b) n > e^{(50)^{\frac{1}{8}}}

    c) n > e^{(100)^{\frac{1}{8}}}

    d) n > e^{(25)^{\frac{1}{8}}}

    e) n > e^{(100)^{\frac{1}{4}}}
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  2. #2
    MHF Contributor matheagle's Avatar
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    integral test, after you let u=\ln x you will get a p-integral.
    You can prove that \sum {1\over n(\ln n)^p} converges whenever p exceeds one.
    Now we need to be careul where we start this series, that's why I left out the n=2....
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    integral test, after you let u=\ln x
    I know, but will that give me the answer? I just wasn't sure how to test it for a specific value.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I only proved it converged.
    To estimate the error, well I guess you should split the sum into two parts.
    One from 2 to N and the other from N+1 to infinity.
    You can then bound that second sum (which is the remainder via the integral)
    \sum_{n=2}^{\infty} {4\over n(\ln n)^9}=\sum_{n=2}^N {4\over n(\ln n)^9}+<br />
\sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}.

    Call this

    \sum_{n=2}^{\infty} {4\over n(\ln n)^9}=S_N+R_N

    You want a bound on R_N=\sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}.

    And you can play the same game with the boxes and use the integral

    \int_N^{\infty} {4dx\over x(\ln x)^9}={4\over 8(\ln N)^8}<.01 , solve for N, giving me (b).

    The you can sum up to that point.
    Last edited by matheagle; March 24th 2009 at 09:09 PM.
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