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

In summary, The conversation discusses the use of the ratio test to determine the convergence of the series \sum\frac{1}{log(n)}. One person attempted to solve it using the ratio test and another suggested using a comparison test instead. The discussion also touches on the definition of the ratio test and its limitations. In conclusion, it is concluded that the ratio test cannot determine the convergence of a series if the limit is exactly 1.
  • #1
DKnight768
4
0
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 [tex]\sum\frac{1}{log(n)}[/tex] converges.


Homework Equations



[tex]\sum\frac{1}{log(n)}[/tex]


The Attempt at a Solution


Well, I tried [tex]\frac{a_{n+1}}{a_{n}}[/tex] and got [tex]\frac{log(n)}{log(n+1)}[/tex].
I can also try [tex]a_{n}=a_{n-1}[/tex] and then it's [tex]\frac{log(n-1)}{log(n)}[/tex], 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 [tex]\frac{log(n-1)}{log(n)}[/tex] 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...
 
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  • #2
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.
 
  • #3
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?
 
  • #4
DKnight768 said:
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.
 
  • #5
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?
 
  • #6
DKnight768 said:
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.
 
  • #7
I see. Thanks!
 

1. What is the ratio test for convergence of series 1/log(n)?

The ratio test is a method used to determine the convergence or divergence of a series. Specifically, the ratio test for the series 1/log(n) involves taking the limit of the absolute value of the ratio of consecutive terms in the series. If this limit is less than 1, the series converges. If the limit is greater than 1 or does not exist, the series diverges.

2. How do you apply the ratio test to the series 1/log(n)?

To apply the ratio test to the series 1/log(n), we first take the absolute value of the ratio of consecutive terms, which is (1/log(n+1))/(1/log(n)) = log(n)/log(n+1). Then, we take the limit of this expression as n approaches infinity. If the limit is less than 1, the series converges. If the limit is greater than 1 or does not exist, the series diverges.

3. What is the significance of the ratio test for the series 1/log(n)?

The ratio test is significant because it allows us to determine the convergence or divergence of a series based on the behavior of the terms in the series. In the case of the series 1/log(n), which is a harmonic series with a logarithmic term, the ratio test shows that the series converges despite having infinitely many terms.

4. Are there any other tests that can be used to determine the convergence of the series 1/log(n)?

Yes, there are other tests that can be used to determine the convergence of the series 1/log(n). These include the integral test, comparison test, and limit comparison test. However, the ratio test is often the most efficient and straightforward method for this series.

5. Can the ratio test be used for all series?

No, the ratio test cannot be used for all series. It can only be applied to series where the terms involve powers of n, exponential functions, or factorials. Additionally, the series must have positive terms for the ratio test to be applicable.

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