anonymous
  • anonymous
Really stumped on this one, anyone good at divergence of series?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
hey there is a of getting a bigger picture it is a little a hard to read for me at least just asking..
anonymous
  • anonymous
\[\sum_{3}^{∞} \frac{ 1 }{ n \ln n \sqrt{(\ln n)^2-1} }\]

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abb0t
  • abb0t
It 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
  • anonymous
any suggestions for what to compare it with? I redid the equation for readability
anonymous
  • anonymous
one sec
anonymous
  • anonymous
try this website http://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx
abb0t
  • abb0t
Use integral test.
abb0t
  • abb0t
@godorovg this is for series, not functions in vector fields.
anonymous
  • anonymous
sorry it has been a while for me. I am really sorry man..
abb0t
  • abb0t
You 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 u-substitution.
anonymous
  • anonymous
I am trying to find the right way to help here http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx
anonymous
  • anonymous
is this what I need to look at? I am making sure I do this right is all..
abb0t
  • abb0t
If 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
  • anonymous
okay let me review this
anonymous
  • anonymous
The 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
  • anonymous
that is what I was think also it must be + not - for this test to work as well
anonymous
  • anonymous
according to what I have read on the website at least.
anonymous
  • anonymous
yeah, back to square one I guess
anonymous
  • anonymous
no 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
  • anonymous
also given that when infinty goes to 3 zero would not be included right?
anonymous
  • anonymous
so 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
  • anonymous
Am I making sense to you?
anonymous
  • anonymous
because the limit would be 3 in this case
anonymous
  • anonymous
you still with me??
anonymous
  • anonymous
so let me make sure I do this right, the integral is lim n->inf sec^-1(lnx) evaluated from 3 to t right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
if you come up with zero at any time test fails..
anonymous
  • anonymous
can you check the answer after you are finished?
anonymous
  • anonymous
i got pi/2 -arcsec(ln3), implying that the sum converges
anonymous
  • anonymous
that looks right
terenzreignz
  • terenzreignz
I 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
  • anonymous
did 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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