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Remainder: Alternate Series

  1. Oct 1, 2006 #1
    How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

    [tex] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} [/tex], [tex] | error | < 0.01 [/tex]. So, [tex] b_{n} = \frac{1}{n^{2}} [/tex]. [tex] b_{n} < b_{n+1} [/tex], and [tex] \lim_{n\rightarrow \infty} b_{n} = 0 [/tex]. Therefore, the series is convergent. I wrote out some terms of the series: [tex] s = 1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+\frac{1}{49}-\frac{1}{64}+\frac{1}{81}-\frac{1}{100}+\frac{1}{121}+ . . . [/tex]. From this step, how do we determine the number of terms we need to add in order to find the sum to the indicated accuracy?

    Thanks
     
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  3. Oct 1, 2006 #2

    Hurkyl

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    Well, the easy answer is if |b_n| is decreasing, then the sum of the entire series lies between any two consecutive partial sums.
     
  4. Oct 1, 2006 #3
    oh ok, so just look at the difference between a [tex] b_{n+1}-b_{n} [/tex] and see if it is less than [tex] 0.01 [/tex]? And [tex] n [/tex] is the number of terms you need?
     
  5. Oct 1, 2006 #4

    Hurkyl

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    That's the basic idea. (You wanted absolute value, BTW) I don't think there's an off-by-one error either.
     
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