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Approximating the "Tail" of a Series

  1. Apr 5, 2008 #1
    [SOLVED] Approximating the "Tail" of a Series

    1. The problem statement, all variables and given/known data
    I need to estimate the tail, which is

    [tex]\Sigma[/tex] (from n=6 to infinity) (4-sin n)/(n^2+1)

    It says to do this with an appropriate improper integral or geometric series.

    3. The attempt at a solution

    I don't see a geometric series helping here, so I would use an improper integral. It is too difficult to integrate (4-sin n)/(n^2+1) (if it's even possible), so I'm stuck here. If this can be integrated, please show me the way. I was thinking perhaps do an improper integral for series of 5/n^2 instead since that is always larger (i.e. if you write 4+/- 1)/(n^2+1) instead. Would this be appropriate?
  2. jcsd
  3. Apr 6, 2008 #2

    Gib Z

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    Homework Helper

    Well yes but I would rather do the improper integral for [tex]\frac{4}{n^2+1}[/tex] because the sine will keep alternating between positive and negative, roughly canceling out each others effects, and also why neglect the +1 in the denominator if we don't have to? Integrating is simple now. We can see that when taking the improper integral, the absolute maximum error extracting the sine is about 0.165, since the sine function has a upper bound of 1, so the maximum error would be [tex]\int^{\infty}_6 \frac{1}{x^2+1} dx[/tex] which is about 0.165. In reality it is much less sine the sine alternates between positive and negative, its less than 0.03.

    However, if we were to ignore the +1 in the denominator as well, we would get [tex]4\sum_{n=6}^{\infty} \frac{1}{n^2}[/tex] which is simple if you know [tex]\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}[/tex]. This method gets you about 0.725..
  4. Apr 6, 2008 #3
  5. Apr 10, 2008 #4
    how exactly do you calculate this error? I understand everythign else such as choice of improper integral, but I don't get where your getting the numbers for the error, is it strictly the value of the improper integral?
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