Using the integral test, how do you show whether #sum (1/n^2)cos(1/n) # diverges or converges from n=1 to infinity?

2 Answers
George C.
Sep 14, 2015

The #cos(1/n)# factor is a bit of a distraction.

#sum_(n=1)^oo 1/(n^2)# is absolutely convergent (as can be easily shown by a variety of means) and #abs(cos(1/n)) <= 1#.

Explanation:

Note that #int_1^oo 1/(x^2) dx = -1/x |_1^oo = 0 - (-1) = 1#

So by the integral test #sum_(n=1)^oo (1/(n^2))# is convergent

Now #1/(n^2) > 0# for all #n > 0#, so #sum_(n=1)^oo abs(1/(n^2))# is convergent.

#abs(cos(1/x)) <= 1 AA x > 0#

Therefore:

#sum_(n=1)^oo abs(1/(n^2) cos(1/n)) <= sum_(n=1)^oo abs(1/(n^2))#

is a sum of non-negative terms, bounded above and therefore convergent.

Hence:

#abs(sum_(n=1)^oo (1/(n^2)) cos(1/n)) <= sum_(n=1)^oo abs(1/(n^2) cos(1/n))#

is convergent.

Jim H
Sep 14, 2015

If you insist on using the integral test, you need to find

#int_1^oo 1/x^2 cos(1/x) dx = lim_(brarroo)int_1^b 1/x^2 cos(1/x) dx #
.
Use substitution with #u = 1/x# to get #du = -1/x^2 dx# so the integral becomes:
#lim_(brarroo) -int_1^(1/b)sinu du = lim_(brarroo) -sin(1/b)-(-sin(1)) = sin1#

The integral converges, so the series also converges.

(I think it is obvious that the function #1/x^2 cos (1/x)# is eventually decreasing. Although your grader might want you to show it.)

Related questions
See all questions in Integral Test for Convergence of an Infinite Series
Impact of this question
9484 views around the world
You can reuse this answer
Creative Commons License