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anonymous
 3 years ago
Really stumped on this one, anyone good at divergence of series?
anonymous
 3 years ago
Really stumped on this one, anyone good at divergence of series?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey there is a of getting a bigger picture it is a little a hard to read for me at least just asking..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{3}^{∞} \frac{ 1 }{ n \ln n \sqrt{(\ln n)^21} }\]

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0It is kinda small, but it looks like you'd want to use comparison test. I think you might be able to use integral test also since you are workingn with log, and you know the values must be positive.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0any suggestions for what to compare it with? I redid the equation for readability

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try this website http://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0@godorovg this is for series, not functions in vector fields.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry it has been a while for me. I am really sorry man..

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0You can use to evaluate the series just as you would an improper integral. Try it to see what you get. you are going to be using usubstitution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am trying to find the right way to help here http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this what I need to look at? I am making sure I do this right is all..

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0If you are lookng at that site: http://tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx Example one would be a similar example to the problem you are working on now.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay let me review this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The thing that bothers me, is for the integral test to apply, the derivative must be less than 0, which it doesn't appear to be

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is what I was think also it must be + not  for this test to work as well

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0according to what I have read on the website at least.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, back to square one I guess

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no it does work here because when you look the rule of squares and taking LN that is squared would be at 1 at least not zero.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0also given that when infinty goes to 3 zero would not be included right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so I am thinking the series it is itself would 1,2,3 and so on leaving out and the end result than would be postive so therefore test works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Am I making sense to you?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because the limit would be 3 in this case

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so let me make sure I do this right, the integral is lim n>inf sec^1(lnx) evaluated from 3 to t right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you come up with zero at any time test fails..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you check the answer after you are finished?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got pi/2 arcsec(ln3), implying that the sum converges

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I realise that this might be moot, but how about checking the convergence of \[\frac{1}{n(\ln n)^2}\] instead? Then it can be shown to be convergent, and by the limit comparison test, you'll be able to show that your original series was also convergent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you check the answer or can even do that depending if this an odd or even in the book you are working out of? So, how are you doing so far??
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