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Math Help - dont understand how to approxiamte sum of a series?

  1. #1
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    dont understand how to approxiamte sum of a series?

    it syas approxiamate the sum of the series to four decimal places.. do i just keep on testing out numbers and then taking the average or what?
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  2. #2
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    What is the series?
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  3. #3
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    What is the Summation?
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  4. #4
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    it is ((-1)^n)n/7^n
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  5. #5
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    \displaystyle \sum n \left(\frac{-1}{7}\right)^n n=1 to infinity?
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  6. #6
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    its (-1)^n times n/(7^n)
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  7. #7
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    \displaystyle\sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n n}}<br />
{{7^n }}}
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  8. #8
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    ya thats it.. i just dont understand how to do this. do i set n to a certain value like 5?
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  9. #9
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    EDIT

    Given that a_n\ge a_{n+1}>0~\&~(a_n)\to 0 then \sum\limits_{n = 1}^\infty  {( - 1)^n a_n } converges.

    If S=\sum\limits_{n = 1}^\infty  {( - 1)^n a_n } then \left| {S - \sum\limits_{n = 1}^N {( - 1)^n a_n } } \right| \leqslant a_{N + 1} .
    Last edited by Plato; November 22nd 2010 at 02:44 PM.
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  10. #10
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    Since this is an alternating series, the sum to "N+2" will always lie between the sum to "N" and "N+1". That means that the full sum will lie between the sum to N and N+1. And that means that the error will be no larger than the absolute value of the difference between those two sums which is just the absolute value of the N+ 1 term:
    \frac{N+1}{7^{N+1}}. (I see now that is what Plato just said.)

    In order to get the sum correct within 4 decimal places, that is, with error between -0.00005 and 0.00005, you must use N such that \frac{N+1}{7^{N+1}}< 0.00005
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