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Math Help - Error in alternating series

  1. #1
    Senior Member mollymcf2009's Avatar
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    Error in alternating series

    I really need someone to walk me through this step by step. I have no idea how to go about this.

    Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

    \sum^{\infty}_{n=1}  \frac{(-1)^{n+1}}{7n^4}

    (|error|) < .00005
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  2. #2
    MHF Contributor matheagle's Avatar
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    In an alternating series with the absolute value of the terms decreasing towards zero we have...

     \biggl|S_n-L\biggr|<a_{n+1}

    where L is the value of the entire sum, S_n is the n  ^{th} partial sum and a_{n+1} is the next term.
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  3. #3
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    skeeter's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    I really need someone to walk me through this step by step. I have no idea how to go about this.

    Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

    \sum^{\infty}_{n=1}  \frac{(-1)^{n+1}}{7n^4}

    (|error|) < .00005
    note that the basic series is an alternating p-series, p > 1, so convergence is absolute.

    error for an alternating series is less than the first omitted term ...

    \left|\frac{(-1)^n}{7n^4}\right| < .00005

    \frac{1}{7n^4} < .00005

    7n^4 > \frac{1}{.00005} = 20000

    7 \cdot 7^4 = 7^5 = 16807

    7 \cdot 8^4 = 28672


    n = 7 should do it
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Thank you!!!

    Thanks Skeeter! That is EXACTLY what I was looking for!!
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