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By far, these infinite series and sequences have been my biggest problem in calculus 2, and this problem really puts my limited knowledge to the test.

I have found that posting these pictures helps show everyone what I have learned so far, and then I'll try to solve it myself.

Chapter and section names:

enter image description here

Rules that my book gives me to solve these problems. This shows more of what I should and do know:

enter image description here

Looking at those rules, it doesn't really even seem to help me with this problem. The only way I could see this being solved with what I know is with the integral test.

Also, note that it starts at 2 (I almost missed that myself).

If I'm going to try to use the integral test, then here is my book's definition of it:

enter image description here

And here is the question again: Determining if the series $\sum_{n=2}^\infty {1\over {nln(n)ln(ln(n))}}$ is convergent or divergent and justifying the answer. I honestly don't know where to start with this one, I would try but there are a lot of changes from previous questions, and it makes it a lot harder.

Any help is greatly appreciated. Thank you.

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marked as duplicate by Nosrati, Daniel Fischer calculus Oct 28 '17 at 20:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know the integral of $\frac{1}{n\ln n\ln (\ln n) }$? $\endgroup$ – kingW3 Oct 28 '17 at 20:31
  • $\begingroup$ Cauchy condensation test. $\endgroup$ – user228113 Oct 28 '17 at 20:31
  • $\begingroup$ @G.Sassatelli I have not learned about that yet. $\endgroup$ – JustHeavy Oct 28 '17 at 20:32
  • $\begingroup$ how about let $x=ln(n)$? $\endgroup$ – valer Oct 28 '17 at 20:34
  • $\begingroup$ @kingW3 I came up with $ln(\left\lvert {ln(ln(x))} \right\lvert) + C$ $\endgroup$ – JustHeavy Oct 28 '17 at 20:35
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This is the Cauchy condensation test, look at $\sum 2^n a_{2^n}$. This gives $$\sum \frac{1}{n\ln n}$$ and applying it a second time we get $$\sum \frac{1}{n}$$ which diverges, thus the above two series also diverge.

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  • $\begingroup$ I have not learned about this test yet. But if I were to use the integral test do you think it would work? I'm having trouble making sure it fits the parameters that my book gives me (I posted my book's definition above). $\endgroup$ – JustHeavy Oct 28 '17 at 20:50
  • $\begingroup$ Its too hard to integrate. $\endgroup$ – Rene Schipperus Oct 28 '17 at 20:52
  • $\begingroup$ Hmmm, ok, thank you for trying to help. I'll have to ask for help outside of StackExchange. $\endgroup$ – JustHeavy Oct 28 '17 at 20:54

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