Sum of 1/ln(n)^(ln(n)) from n=3 to infinity?

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Homework Help Overview

The discussion revolves around the convergence or divergence of the series defined by the sum of 1/ln(n)^(ln(n)) from n=3 to infinity, a topic situated within the realm of series convergence in mathematical analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the convergence of the series and exploring the relationship between ln(n)^(ln(n)) and n^(ln(ln(n))). There is an attempt to clarify this relationship through mathematical manipulation.

Discussion Status

The discussion is ongoing, with participants engaging in attempts to demonstrate the equivalence of ln(n)^(ln(n)) and n^(ln(ln(n))). Some guidance has been offered regarding the use of exponential and logarithmic properties, but no consensus has been reached on the convergence of the series.

Contextual Notes

There appears to be a playful tone in the exchanges, with some participants expressing surprise at the mathematical manipulations being discussed. The original poster's inquiry about convergence remains central to the discussion.

sedaw
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Sum of 1/ln(n)^(ln(n)) from n=3 to infinity?
Does the sum of 1/ln(n)^(ln(n)) from n=3 to infinity converge or diverge?

TNX ...
 
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Can you show ln(n)^(ln(n))=n^(ln(ln(n)))?
 


Dick said:
Can you show ln(n)^(ln(n))=n^(ln(ln(n)))?

nope , you are magician , how u did it ?


TNX !
 


sedaw said:
nope , you are magician , how u did it ?


TNX !

Try it doing it. Use that a^b=e^(ln(a)*b).
 

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