What function should I use for comparison

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Homework Help Overview

The discussion revolves around the convergence of the series ln(k)/k^2 and the appropriate comparison series to use for analysis. Participants are exploring the implications of the logarithmic term in the numerator and its effect on convergence behavior.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the suitability of comparing ln(k)/k^2 to 1/k^2, questioning the implications of the logarithmic term. There is an exploration of whether the series converges or diverges, and suggestions for alternative comparison series are raised.

Discussion Status

Some participants have offered hints and suggestions for potential comparison series, while others are considering the implications of their assumptions about convergence. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

Participants are navigating the constraints of using the comparison test and are considering the nature of the series involved, including the behavior of ln(k) as k approaches infinity.

armolinasf
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Homework Statement



If i have the series ln(k)/k^2 would I compare it to 1/k^2? The reason why I'm confused is that ln(k)/k^2 > 1/k^2 so if its greater than a function that converges it doesn't tell us anything right? So I'm not sure what exactly to compare it with on account of the ln(k) in the numerator. thanks for the help.
 
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hello,

you are right that 1/k^2 will not help us in this case.

the first thing you need to decide is whether you think your series converges or diverges; that will determine the kind of series you want to compare it to. what do you think?
 
my guess would be that it converges
 
ok, then we need a series that converges, but that is bigger than ours.

do you have to use the comparison test, or can yo use any method?
 
what series have you tried?
 
Here's a hint: k^(1/2) > ln(k) for all k.
 
Alright if k^1/2 is greater than ln k i could compare it with K^(1/2)/k^2 = 1/k^3/2 which converges since p>1.

One more question: what if it were 1/ln(k)?
 
good!

well, for 1/ln(k) we again first need to decide if we think it converges or diverges. this one should be pretty easy if we think carefully. if we think it diverges (and hopefully we do :smile:), we need f(k) > ln(k) and 1/f(k) divergent. that shouldn't be too hard to find.

cheers
 

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