(-1)^(n+1) ln x /n^2 tell if converge or diverge?

Question

(-1)^(n+1) ln x /n^2 tell if converge conditionally, converge absolutely or diverge

I have been trying to use all the rules an tests to see if ln n / n^2 is convergent or divergent and can't get it.
I think the limit of it =0 so nth term test doesn't work. The ratio test didn't work for me either. Can I use the p series test? or the nth root test?

If it does converge how do you know if it is conditonally or absolutely?

Answer 1

If you could verify that ln n <= n^1/2 (even for just sufficiently large n, and not for all) then that series will be convergent.

all I know is ln n <= n (for large n) then that makes the series conditionally convergent at least. (Thus so far, it is not divergent.)

My 'question' is really the determination whether it is conditional or absolute. d:

Edit:
Okay, this is one approach I can think. If n^(1/2)/ln(n) approaches infinity then the numerator is undoubtedly greater than the denominator.

Getting the limit by L'Hopitals rule is easy. The limit is infinity.
Thus, for sufficiently large n,

ln n/ n^2 < n^(1/2)/n^2 = 1/n^(3/2), the series of whom is absolutely convergent.

Then the series ln n/ n^2 is also absolutely convergent.
Then the series (-1)^(n+1) ln n/ n^2 is absolutely convergent.
:)

Answer 2

Taken in absolute value the series is (n+1)/n^2 *I ln xI of the same class that IlnxI/n divergent unless x=1 when a_n=0
So the series could be only conditionally convergent
Let´s see if it is decreasing monotonously

Take z = (y+1)/y^2 Iln xI as a function of y
z´= 1/y^4(y^2-y^2-y) *IlnxI = -1/y^3 *IlnxI <0
so (n+1)/n^2 Iln xI is monotonously decreasing to zero and the series (alternate) is conditionally convergent