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Math Help - Determining absolute/conditional convergence, approximating sum

  1. #1
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    Determining absolute/conditional convergence, approximating sum

    Determine if \sum_{n=0}^{\infty} \frac {(-1)^{n+1}}{n^5+n+6} converges absolutely or conditionally. Approximate the sum within an accuracy of 0.01.

    \sum_{n=0}^{\infty} |\frac {(-1)^{n+1}}{n^5+n+6}|
    \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}
    Basic comparison test with \sum \frac {1} {n^5}.
    \sum \frac {1} {n^5} converges, so \sum_{n=0}^{\infty} \frac {1}{n^5+n+6} converges as well. The series converges absolutely.

    I'm having trouble understanding how to approximate. Do I set up random partial sums until it's less than 0.01? How do I know how how many terms I need for it to be less than 0.01? Is there a formula?
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  2. #2
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    Re: Determining absolute/conditional convergence, approximating sum

    Quote Originally Posted by deezy View Post
    Determine if \sum_{n=0}^{\infty} \frac {(-1)^{n+1}}{n^5+n+6} converges absolutely or conditionally. Approximate the sum within an accuracy of 0.01.

    \sum_{n=0}^{\infty} |\frac {(-1)^{n+1}}{n^5+n+6}|
    \sum_{n=0}^{\infty} \frac {1}{n^5+n+6}
    Basic comparison test with \sum \frac {1} {n^5}.
    \sum \frac {1} {n^5} converges, so \sum_{n=0}^{\infty} \frac {1}{n^5+n+6} converges as well. The series converges absolutely.

    I'm having trouble understanding how to approximate. Do I set up random partial sums until it's less than 0.01? How do I know how how many terms I need for it to be less than 0.01? Is there a formula?
    since the series is alternating , error < |first omitted term|

    Error < \frac{1}{100} \le \frac{1}{n^5+n+6}

    looks like n = 3 will get you there

    so ...

    \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^5+n+6} \approx -\frac{1}{6} + \frac{1}{8} - \frac{1}{40} will be w/in the desired error bound
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