Question

Really stumped on this one, anyone good at divergence of series?

Answer 1

hey there is a of getting a bigger picture it is a little a hard to read for me at least just asking..

Answer 2

\[\sum_{3}^{∞} \frac{ 1 }{ n \ln n \sqrt{(\ln n)^2-1} }\]

Answer 3

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.

Answer 4

any suggestions for what to compare it with? I redid the equation for readability

Answer 5

one sec

Answer 6

try this website http://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx

Answer 7

Use integral test.

Answer 8

@godorovg this is for series, not functions in vector fields.

Answer 9

sorry it has been a while for me. I am really sorry man..

Answer 10

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.

Answer 11

I am trying to find the right way to help here
http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

Answer 12

is this what I need to look at? I am making sure I do this right is all..

Answer 13

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.

Answer 14

okay let me review this

Answer 15

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

Answer 16

that is what I was think also it must be + not - for this test to work as well

Answer 17

according to what I have read on the website at least.

Answer 18

yeah, back to square one I guess

Answer 19

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.

Answer 20

also given that when infinty goes to 3 zero would not be included right?

Answer 21

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

Answer 22

Am I making sense to you?

Answer 23

because the limit would be 3 in this case

Answer 24

you still with me??

Answer 25

so let me make sure I do this right, the integral is lim n->inf sec^-1(lnx) evaluated from 3 to t right?

Answer 26

yes

Answer 27

if you come up with zero at any time test fails..

Answer 28

can you check the answer after you are finished?

Answer 29

i got pi/2 -arcsec(ln3), implying that the sum converges

Answer 30

that looks right

Answer 31

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.

Answer 32

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??