Convergent or Divegent Series?

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Homework Help Overview

The problem involves the convergence or divergence of the series \(\sum{(ln(k))/(\sqrt{k+2})}\), with \(k\) starting at 1 and extending to infinity. Participants are exploring the applicability of various convergence tests and the behavior of the series as \(k\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of L'Hopital's rule and the limit of individual terms, questioning its relevance to series convergence. There is a suggestion to compare the series to \(\frac{1}{\sqrt{k+2}}\) for easier analysis of convergence.

Discussion Status

Some participants are clarifying the importance of considering partial sums in the context of series convergence. There is an acknowledgment of the relationship between the original series and the comparison series, indicating a productive direction in the discussion.

Contextual Notes

Participants note the need to understand the definition of series convergence and the implications of comparing terms in the series. There is an ongoing exploration of assumptions regarding the behavior of the terms as \(k\) increases.

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Homework Statement


[tex]\sum{(ln(k))/(\sqrt{k+2})}[/tex], with k starting at 1 and going to [tex]\infty[/tex]


Homework Equations


Does this series converge or diverge? Be sure to explain what tests were used and why they are applicable.


The Attempt at a Solution


Okay, my TA got that this diverges, but I got that it converges by simply taking the limit as k goes to [tex]\infty[/tex] and applying L'Hopital's rule. I also plugged the function into my calculator and it seems to converge at y=0, which is corroborates what I got with l'hospital's rule. What did you guys get? Any help is greatly appreciated.
 
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It looks like you proved the individual terms go to 0, which isn't what you're trying to do. The definition of a series is you take the limit of the partial sums. And where does y come into the series?

Try comparing the series to [tex]\frac{1}{\sqrt{k+2}}[/tex] whose convergence/divergence is easier to find
 
Right, I forgot about the partial sums. Thanks, that helped a lot. But isn't [tex]1/(\sqrt{k+2})[/tex] smaller than [tex](ln(k))/(\sqrt{k+2})[/tex]?
 
Last edited:
iatnogpitw said:
Right, I forgot about the partial sums. Thanks, that helped a lot. But isn't [tex]1/(\sqrt{k+2})[/tex] smaller than [tex](ln(k))/(\sqrt{k+2})[/tex]?
Yeah, it is. You've been handed a clue for free. If you can say something about what [tex]\sum 1/(\sqrt{k+2})[/tex] does, then maybe you will know something about the series you're really interested in.
 

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