Determine convergence for a series

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    Convergence Series
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Homework Help Overview

The discussion revolves around determining the convergence of the series involving the term ln(n)/(ln(ln(n))) for n starting from 3 to infinity. Participants express uncertainty about the appropriate comparison series to use for applying the limit test.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants consider using the limit comparison test with the series 1/n. There is discussion about whether this comparison is valid and what justifications might be necessary. Some participants also suggest that the behavior of log(n) compared to log(log(n)) could inform their approach.

Discussion Status

Participants are actively exploring different methods to assess the convergence of the series. Some have suggested using the nth term test for divergence based on observations about the growth rates of the logarithmic functions involved. There is no explicit consensus yet, but several lines of reasoning are being examined.

Contextual Notes

Participants note that the tail of the series does not seem to approach zero, which raises questions about the validity of the series' convergence. There is also mention of the need to justify comparisons made in the limit test.

MozAngeles
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Homework Statement


Sum ln n/(ln(ln n)) n=3..infinity?
Im pretty sure it diverges and I am pretty sure to use the limit test but i just don't know what to compare this sum to. Would 1/n be ok. Do i have to justify why they are similar?

ANy help would be nice thanks.


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The Attempt at a Solution

 
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MozAngeles said:

Homework Statement


Sum ln n/(ln(ln n)) n=3..infinity?
Im pretty sure it diverges and I am pretty sure to use the limit test but i just don't know what to compare this sum to. Would 1/n be ok. Do i have to justify why they are similar?

ANy help would be nice thanks.
Sure, you can use 1/n in the limit comparison test.

If this limit is positive, then your series diverges.
[tex]\lim_{n \to \infty}\frac{\frac{ln(n)}{ln(ln(n))}}{1/n}[/tex]
 
You could also just see that log(n) is larger than log(log(n)) for basically every n you're using, and so you're summing a sequence of numbers that all are greater than 1. So realistically, you could use a limit comparison with 1.
 
even worse, the tail of this series doesn't seem to go to 0, log(x) will grow faster than log(log(x))
 
Char. Limit said:
You could also just see that log(n) is larger than log(log(n)) for basically every n you're using, and so you're summing a sequence of numbers that all are greater than 1. So realistically, you could use a limit comparison with 1.

wisvuze said:
even worse, the tail of this series doesn't seem to go to 0, log(x) will grow faster than log(log(x))

These observations suggest that the nth term test for divergence might be appropriate here.
 
thanks i figured it out like two seconds after i posted. haha
 

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