Series Test for convergent and divergent

[mikbear]

Homework Statement

Ʃ √n/(ln(n))^2
from n=2 to ∞

Homework Equations

Series Test for convergent and divergent

The Attempt at a Solution

I tried doing ratio test and gotten
[√(n+1)*(ln(n))^n] / [(ln(n+1))^(n+1) * √n]

to find the limit, do i cont by using Hopstal rule?

 

[Mark44]

Homework Statement

Ʃ √n/(ln(n))^2
from n=2 to ∞

Homework Equations

Series Test for convergent and divergent

The Attempt at a Solution

I tried doing ratio test and gotten
[√(n+1)*(ln(n))^n] / [(ln(n+1))^(n+1) * √n]

That is not correct. It should be
$$ \frac{\sqrt{n+1}~(ln(n))^2}{\sqrt{n}~(ln(n+1))^2}$$

In any case, I don't think the Ratio Test is going to be much help here. What other tests do you know?

to find the limit, do i cont by using Hopstal rule?

 

[Dick]

The absolute first check you should always make is to check that the nth term approaches 0. l'Hopital's should help there.

 

[mikbear]

I made a mistake in the question its suppose to be power of n

Ʃ √n/(ln(n))^n
from n=2 to ∞

I have learned root test, integral test, comparison and limit test.

however I do not see how these will help solve this question.

 

[Dick]

I made a mistake in the question its suppose to be power of n

Ʃ √n/(ln(n))^n
from n=2 to ∞

I have learned root test, integral test, comparison and limit test.

however I do not see how these will help solve this question.

That's little more subtle. Can you show (ln(n))^n will eventually dominate any power series? For example, (ln(n))^n>n^2 for sufficiently large n? That would let you make a comparison test.

 

[mikbear]

Oh. thanks for the tip. I gona try it rite now. Thanks