## ieatpi 2 years ago Really stumped on this one, anyone good at divergence of series?

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1. ieatpi

2. godorovg

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

3. ieatpi

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

4. 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.

5. ieatpi

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

6. godorovg

one sec

7. godorovg
8. abb0t

Use integral test.

9. abb0t

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

10. godorovg

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

11. 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.

12. godorovg

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

13. godorovg

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

14. 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.

15. godorovg

okay let me review this

16. ieatpi

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

17. godorovg

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

18. godorovg

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

19. ieatpi

yeah, back to square one I guess

20. godorovg

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.

21. godorovg

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

22. godorovg

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

23. godorovg

Am I making sense to you?

24. godorovg

because the limit would be 3 in this case

25. godorovg

you still with me??

26. ieatpi

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

27. godorovg

yes

28. godorovg

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

29. godorovg

can you check the answer after you are finished?

30. ieatpi

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

31. godorovg

that looks right

32. 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.

33. godorovg

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??