Ratio test for convergence of series 1/log(n)?

[DKnight768]

Hi. I have been banging my head against this problem for a while and I just don't get it. Maybe (probably) it's something wrong with my logarithm-fu or limit-fu. I just registered to ask this because I couldn't find an answer anywhere, and I've been reading these forums for a while for other problems and they've been a big help.

Homework Statement

Use the ratio test to find if \sum\frac{1}{log(n)} converges.

Homework Equations

\sum\frac{1}{log(n)}

The Attempt at a Solution

Well, I tried \frac{a_{n+1}}{a_{n}} and got \frac{log(n)}{log(n+1)}.
I can also try a_{n}=a_{n-1} and then it's \frac{log(n-1)}{log(n)}, isn't it?
Maybe this is stupid, but can I remove the log() signs from the fraction? I can't, right?
I guess my problem is that I don't know how to compute the limit of \frac{log(n-1)}{log(n)} as n becomes very large. log(n-1) is less than log(n), so maybe it's 0, and since the condition is that c is 0<c<1 it does not converge? I'm not confident at all in that answer. Please help

PS: I hope the latex symbols show up allright...

 

[Dick]

The ratio test won't tell you anything for that series. The limiting ratio is 1. You can use l'Hopital to verify that. Try a comparison test.

 

[DKnight768]

I tried a comparison test and yes, I can solve it with that. But the problem is inside the ratio test problem set, so it should be solvable with a ratio test, or at least mixing ratio and comparison, going by how the book is structured (book is "A First Course in Calculus" by Serge Lang). This is about series only with positive terms by the way.
You're right, using l'hôpital I get 1, can't I conclude then than since the limit c=1 is not less than 1, it does not converge by the definition of the ratio test?

 

[Dick]

I tried a comparison test and yes, I can solve it with that. But the problem is inside the ratio test problem set, so it should be solvable with a ratio test, or at least mixing ratio and comparison, going by how the book is structured (book is "A First Course in Calculus" by Serge Lang). This is about series only with positive terms by the way.
You're right, using l'hôpital I get 1, can't I conclude then than since the limit c=1 is not less than 1, it does not converge by the definition of the ratio test?

Apply the ratio test to 1/n^2. The ratio is also 1. But that converges by an integral test. No, c=1 doesn't tell you diverges. I'm not sure why it's in the ratio test section either.

 

[DKnight768]

Thanks. Last question. So then, if the limit c of the ratio test isn't between 0 and 1, I can't say for sure if it converges or diverges, but if it is between 0 and 1 it converges for sure? did I got that right?

 

[Dick]

Thanks. Last question. So then, if the limit c of the ratio test isn't between 0 and 1, I can't say for sure if it converges or diverges, but if it is between 0 and 1 it converges for sure? did I got that right?

If it's less than 1 you know it converges, if it's greater than 1 you know it diverges. If it's exactly 1 you don't know anything about convergence.

 

[DKnight768]

I see. Thanks!