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Estimating a sum of an Infinite series

  1. Sep 17, 2006 #1

    G01

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    How many terms of :

    [tex] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} [/tex]

    do you have to add to get an error < .01

    Alright, I used the Alternating Series Estimation Theorem since the terms are decreasing and the terms approach 0.

    So, by the theorem, .01 < = [tex] b_{n+1} [/tex] so

    [tex] 1/(n+1)^2 = 1/100[/tex]
    [tex] (n+1)^2 = 100 [/tex]
    [tex] n+1 = 10[/tex]

    So this means that in order to get this error, we have to add 9 terms right? The back of my book says 10 is the answer. Why is that?
     
  2. jcsd
  3. Sep 17, 2006 #2

    0rthodontist

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    You have a less than or equal sign in one place and a less than sign in the other place. You've shown that the error is less than or equal to 1/100 if you evaluate to 9 terms, but you need one more to show that it's actually less than by this method.
     
  4. Sep 17, 2006 #3

    G01

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    AH HAH!!! Thats it!!!!
    Its supposed to be a less than sign, thanks!
     
  5. Sep 18, 2006 #4

    dextercioby

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    BTW, the sum is exactly [itex] \frac{\pi^{2}}{12} [/itex].

    Daniel.
     
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