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.)

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