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Show that the series is convergent

  1. Jul 28, 2012 #1

    knv

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    1. Show that the series is convergent and then find how many terms we need to add in order to find the sum with an error less than .001

    Ʃ (-1)(n-1)/ √(n+3)

    from n = 1---> infinity




    2. I took the derivative.



    3. f(x) = (x+3)-1/2
    f'(x) = -1/2 (x+3)-3/2

    Then I set up the following

    Absolute value (1/(n+1+3)) < .001

    n+4 > (1/.001)2

    Got 999,997 for the answer. not sure what Im doing wrong. Help!

     
  2. jcsd
  3. Jul 28, 2012 #2
    Re: Series

    There really is no need to be taking derivatives here. You should try to apply the Alternating Series Test. If you don't know what that is, you can find it on Wikipedia. The page has all of your answers.
     
  4. Jul 29, 2012 #3

    knv

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    Re: Series

    would it be 999,998 terms ?
     
  5. Jul 29, 2012 #4

    SammyS

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    Re: Series

    How did you get that?

    The absolute value of the 999,997th term is indeed 0.001 .

    But the series is alternating and the absolute value of subsequent terms if decreasing. If we let Sk represent the kth partial sum, then I would expect the series to converge to a value very close to midway between Sk and Sk+1, for large k.
     
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