Question

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

Answer 1

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.

Answer 2

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.)