Integral test - ultimately decreasing function

  • Thread starter TysonM8
  • Start date
  • #1
25
1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,740
412
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"?
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].

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.
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.
 

Related Threads on Integral test - ultimately decreasing function

Replies
8
Views
5K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
5K
Replies
4
Views
2K
Replies
6
Views
3K
  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
Top