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

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

abb0tBest ResponseYou've already chosen the best response.0
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.
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
any suggestions for what to compare it with? I redid the equation for readability
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
try this website http://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
@godorovg this is for series, not functions in vector fields.
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
sorry it has been a while for me. I am really sorry man..
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
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 usubstitution.
 one year ago

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

godorovgBest ResponseYou've already chosen the best response.0
is this what I need to look at? I am making sure I do this right is all..
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
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.
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
okay let me review this
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
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
 one year ago

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

godorovgBest ResponseYou've already chosen the best response.0
according to what I have read on the website at least.
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
yeah, back to square one I guess
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

godorovgBest ResponseYou've already chosen the best response.0
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
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
Am I making sense to you?
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
because the limit would be 3 in this case
 one year ago

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

godorovgBest ResponseYou've already chosen the best response.0
if you come up with zero at any time test fails..
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
can you check the answer after you are finished?
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
i got pi/2 arcsec(ln3), implying that the sum converges
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
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.
 one year ago

godorovgBest ResponseYou've already chosen the best response.0
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??
 one year ago
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