# Does it converge or diverge?

1. Dec 3, 2009

### Monte_Cristo

1. The problem statement, all variables and given/known data

Determine whether the infinite series converge or diverge?

∑ 1/{1+n[(ln n)^2]}
n=0

2. Relevant equations

Direct Comparison test
Limit Comparison test
Root test
Ratio test
integral test

3. The attempt at a solution

Personally, I think that only two methods can work:

1) Manual method - meaning writing down the series and then look at if its converging or diverging.

2) Direct Comparison test - If I compare it directly with the limit of n approaching infinity for 1/n, then my answer comes out to be that the "series diverges

If I use 1/n[(ln n)^2] or else 1/[(ln n)^2] for direct comparison test, I get the same result.

BUT, when I do it the manual way, my answer turns out to be CONVERGENT.

Thanks

Last edited: Dec 3, 2009
2. Dec 3, 2009

### Staff: Mentor

I have no idea of what you mean by the "manual" method.
Maybe or maybe not. Did you show that each term of your series is larger than the corresponding term of the series $\sum 1/n$?
Do you know for a fact that these series diverge?

3. Dec 3, 2009

### Monte_Cristo

oh my goodness! Now I see that why the Direct Comparison Test can not work on this solution.

By saying "manual way", I mean that if I write down the series (a1 + a2 + a3......an), then my answer turns out to be convergent.

Thanks for your help! But it would be good if I can get some more! :) This problem is hard! All I know is that I am NOT supposed to solve this problem in the "Manual way" - I need to use one of the tests to prove it - and I am TOTALLY stuck! :)

Thanks

4. Dec 3, 2009

### Staff: Mentor

But your series is an infinite series, so there is no way you can write all of its terms. They way you have written it -(a1 + a2 + a3......an) - suggests that there is a last term, which is not the case for infinite series.

You listed several tests that you can use. I wouldn't try the integral test or the n-th root test, but I would try the limit comparison and ratio tests, and maybe the direct comparison test. When you use the two comparison tests, you have to have a few series in your pocket to compare against, such as sum(1/n) and sum(1/n^2), two series that respectively diverge and converge.

5. Dec 3, 2009

### Atropos

Are you sure the summation is from 0 to $$\infty$$, because ln(n) doesn't exist for n=0. Maybe it's supposed to be a trick question?

6. Dec 3, 2009

### Monte_Cristo

I deeply apologize, the summation is:

∑ 1/{1+n[(ln n)^2]}
n=1

I am so sorry that I made the typo error. lol but still I can't find the solution! :)

7. Dec 3, 2009

### Dick

Apply the integral test to 1/(n*ln(n)^2). What do you conclude? How does that series compare with 1/(1+n*ln(n)^2)?