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.)

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