Using the integral test, how do you show whether #sum 1/n^3# diverges or converges from n=1 to infinity?

1 Answer
Jim H
Jun 18, 2015

Check that #1/x^3# is (eventually) decreasing, then see whether #int_1^oo 1/(x^3) dx# diverges or converges.

Explanation:

If your grader or teacher won't let you claim that it is obvious that #1/x^3# is a decreasing function on #[1, oo]#, then

point out that for #f(x) = 1/x^3#, we get #f'(x) = -3/x^4# which is always negative, so #f# is always decreasing.

#int_1^oo 1/(x^3) dx = lim_(brarroo) int_1^b 1/(x^3) dx#

# = lim_(brarroo) (-1/(2x^2)]_1^b)#

# = lim_(brarroo) (-1/(2b^2) + 1/2)#

# = 1/2#

The integral converges, therefore the series converges. (We have not found what the series converges to.)

Related questions
See all questions in Integral Test for Convergence of an Infinite Series
Impact of this question
4656 views around the world
You can reuse this answer
Creative Commons License