Question

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

Answer 1

This may be a "trick question". The term `1/(n(lnn)^2)` is not defined for `n = 1`.

Explanation:

I don't know whether this is a trick question the student was asked of if there is an error in posting it here.

If we eliminate the first term and do the integral test for `sum_2^oo 1/(n(lnn)^2)`, then
I think it is fairly clear that the function `f(x) = 1/(x(lnx)^2)` is eventually non-negative and monotone decreasing, so the challenge is to integrate the function on `[1,oo)`

`int_2^oo 1/(x(lnx)^2) dx = lim_(brarroo)int_2^b 1/(x(lnx)^2) dx`

`= lim_(brarroo)int_2^b 1/(lnx)^2 1/x dx`

`= lim_(brarroo)int_2^b (lnx)^-2 1/x dx`

`= lim_(brarroo) [-(lnx)^-1]_2^b`

`= lim_(brarroo) [-1/lnb - (-1/ln2)]`

`= 0+1/ln2`

So the integral and the series both converge.

Reminder

If we really wanted to we could integrate:

`int_1^oo 1/(x(lnx)^2) dx` by evaluating both:

`int_1^c f(x) dx` and `int_c^oo f(x) dx`

`lim_(ararr1^+) int_a^c 1/(x(lnx)^2) dx` `" "` and `" "` `lim_(brarroo) int_c^b 1/(x(lnx)^2) dx`

for `c` in `(1,oo)`

The integral `int_1^c f(x) dx` diverges, so `int_1^oo 1/(x(lnx)^2) dx` diverges.