Convergent or Divegent Series?

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iatnogpitw
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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]?
 
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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.