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Integral test HELP!

  1. Dec 9, 2007 #1
    Use the integral test to determine convergence or divergence of the foloowing series:

    a) [tex]\infty[/tex] (n=1) [tex]\sum[/tex]n/n[tex]\hat{}[/tex]2+1

    b) [tex]\infty[/tex] (n=2) [tex]\sum[/tex]1/n[tex]\hat{}[/tex]2+1
  2. jcsd
  3. Dec 9, 2007 #2


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    What has your approach been?

    Any thoughts/results (even if incomplete)?
  4. Dec 10, 2007 #3
    Hi, i would proceed as follows:

    In the first case, one can easily see that the function f(n) is monotone decreasing and allways non-negative, so the integral test applies in this situation. Then, the series will
    converge only if the integral from 1 to infinity of x/(x^2+1) dx converges (sory about the equations, i don't know how to use latex yet). And this last integral is the antiderivative of (1/2)*ln(x^2+1), so when you calculate the integral you have to evaluate (1/2)*ln(x^2+1) in infinity and substract (1/2)*ln(x^2+1) evaluated in 1. It is easy to see that this integral diverges.

    In the second one, you proceed in the same way, but in this case you have the integral from 2 to infinity of 1/(x^2+1)dx. This is the antiderivative of Arctan(x), so if you evaluate it in x=infinity and in x=2 you have that the integral is Arctan(infinity)-Arctan(2)=Pi/2-Arctan(2) which is of course a finite value, so in this case the sum converges.

    Sory about the englis i don'tknow it eather very well
  5. Dec 10, 2007 #4


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    Huh? You use TeX-tags, but merely to type \hat{} instead of ^ in "n^2" .
    Why not just use them properly, as in:

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

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

    Just a little tip to make your post more clear :smile:
  6. Dec 10, 2007 #5
    the integral test seems like overkill for establishing divergence/convergence of these two series.

    comparison test seems easier.

    n/(n^2+1) > n/(n^2+n) = 1/n+1 whose sum goes to infinity so that (a) diverges to infinity

    1/(n^2+1) < 1/n^2 which converges b/c the infinite sum of 1/n^p converges for all p>1, so (b) converges (Actually to (pi^2)/6, the proof of this is beautiful and is due to Euler)
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