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Integral test - ultimately decreasing function

  1. Sep 16, 2013 #1
    My textbook states that "For the integral test to apply, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing."

    I'm just curious what is meant by this statement. How do you define a function that is "ultimately decreasing"? It'd be great if you could find an example of a function that increases over a certain domain but ultimately decreases, in which the integral test applies.
     
  2. jcsd
  3. Sep 16, 2013 #2

    pasmith

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    A function [itex]f[/itex] is "ultimately decreasing" if there exists [itex]R[/itex] such that [itex]f(x) > f(y)[/itex] whenever [itex]R \leq x < y[/itex].

    Consider
    [tex]\sum_{n = 0}^{\infty} \frac{1}{1 + (n - 5)^2}[/tex]

    [itex]f(x) = (1 + (x - 5)^2)^{-1}[/itex] is decreasing for [itex]x \geq 5[/itex]. The fact that [itex]f[/itex] is increasing for [itex]0 \leq x \leq 5[/itex] does not affect convergence. We have
    [tex]
    \int_0^{\infty} \frac{1}{1 + (x - 5)^2}\,\mathrm{d}x =
    \left[ \arctan(x - 5)\right]_{0}^{\infty} = \frac{\pi}{2} - \arctan(-5) = \frac{\pi}{2} + \arctan(5)[/tex]
    so the series converges.
     
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