# Integral test - ultimately decreasing function

TysonM8
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.

Homework Helper
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 $f$ is "ultimately decreasing" if there exists $R$ such that $f(x) > f(y)$ whenever $R \leq x < y$.

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
$$\sum_{n = 0}^{\infty} \frac{1}{1 + (n - 5)^2}$$

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