Given series $$\sum_{n=2}^{\infty} \frac{1}{n (\ln n)(\ln \ln n)^2} .$$ To determine whether it is convergent or divergent.

I tried with ratio test but it is inconclusive. Cauchy condensation test was also not very helpful. Then I tried to use the fact that $\ln n<n$, then $\ln \ln n<\ln n$ which also does not bring me any closer to the answer. I don't know much of the inequality involving $\ln$ so I am having problems to solve this. How can I do this? Any hint or help will be great.

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    $\begingroup$ Try the integral test; then let $u=\ln \ln x$. $\endgroup$ – Christopher Carl Heckman May 3 '16 at 5:55
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    $\begingroup$ Cauchy Condensation works. Easier is the integral test. $\endgroup$ – André Nicolas May 3 '16 at 5:56
  • $\begingroup$ @AndréNicolas check my answer please. $\endgroup$ – Kushal Bhuyan May 3 '16 at 6:06
  • $\begingroup$ It is basically fine. There is a typo, you wanted to integrate $\frac{1}{(x)(\ln x)(\ln\ln x)^2}$. $\endgroup$ – André Nicolas May 3 '16 at 6:09
  • $\begingroup$ Oh ok got that typo. $\endgroup$ – Kushal Bhuyan May 3 '16 at 6:09

Using integral test, $\int_{2}^{\infty}\frac{1}{x\ln x (\ln \ln x)^2}dx$.

Taking $u=\ln \ln x$, we get $\lim_{b\to \infty}\int_{2}^{b}\frac{dx}{x\ln x(\ln \ln x)^2}=\lim_{b \to \infty}\int_{\ln \ln3}^{\ln \ln b}\frac{du}{u^2}=-\frac{1}{u}|_{\ln \ln 3}^{\ln \ln b}\to \frac{1}{\ln \ln 3}$ as $b\to \infty$. Hence the series converges.

  • $\begingroup$ The bounds for the integral with $u$ should be $\ln \ln 2$ to $\ln \ln b$. Of course, that doesn't change the conclusion. $\endgroup$ – JimmyK4542 May 3 '16 at 6:20
  • $\begingroup$ The lower limit of the $u$ integral needs to be changed. $\endgroup$ – Rick Sanchez May 3 '16 at 6:22
  • $\begingroup$ Since $\log\log 2 < 0$, the interval you integrate over $\frac{1}{u^2}$ contains the non-integrable singularity at $u = 0$. You need to change the lower limit of integral over $x$ to something $> e$. $\endgroup$ – achille hui May 3 '16 at 6:58
  • $\begingroup$ Ok can you suggest me the change? @achillehui $\endgroup$ – Kushal Bhuyan May 3 '16 at 7:00
  • $\begingroup$ Anything strictly larger than $e$ should work, I will just change $2$ to $3$ and handle the $n = 2$ term in your sum separately. $\endgroup$ – achille hui May 3 '16 at 7:02

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