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Could I have a hint for testing the convergence of the following series please?

$$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$

Edit

The integral test does not work because $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ has not an elementary primitive.

Thank You.

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  • $\begingroup$ did you try the integral test? $\endgroup$ – Alex Jan 26 '14 at 19:57
  • $\begingroup$ en.wikipedia.org/wiki/Sophomore's_dream $\endgroup$ – Alex Jan 26 '14 at 20:00
  • $\begingroup$ @Alex I'm trying with the integral test, but this integral is really tough for me... I'll check your link now! Thank you for your help man; I really appreciate it. $\endgroup$ – Charlie Jan 26 '14 at 20:03
  • $\begingroup$ @Alex Am I on the right way? Can this integral be reduced to a simple one? $\endgroup$ – Charlie Jan 26 '14 at 20:41
  • $\begingroup$ Related: math.stackexchange.com/questions/408824/… $\endgroup$ – Hans Lundmark Dec 9 '17 at 14:45
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Alternate hint:

$$ (\ln n)^{\ln n} = n^{\ln \ln n}. $$

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  • $\begingroup$ Thank you. I did it with the comparison test. $\endgroup$ – Charlie Jan 26 '14 at 22:18
  • $\begingroup$ You're welcome :) $\endgroup$ – Antonio Vargas Jan 26 '14 at 23:31
  • $\begingroup$ I have never seen this before, how do you get to that equation? $\endgroup$ – Nhat Jan 27 '14 at 17:20
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    $\begingroup$ @KitKat, take $\ln$ of $(\ln n)^{\ln n}$ to get $(\ln n)\ln\ln n$, rewrite this as $\ln(n^{\ln\ln n})$, then exponentiate this to get $n^{\ln \ln n}$. $\endgroup$ – Antonio Vargas Jan 27 '14 at 17:38
  • $\begingroup$ @AntonioVargas Nice, thanks. $\endgroup$ – Nhat Jan 27 '14 at 19:52
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Hint:

Consider

$$ 2^k \frac{1}{(k\ln e)^{k\ln 2}}>\sum_{2^k\le n< 2^{k+1}} \frac{1}{(\ln n)^{\ln n}}> 2^k\frac{1}{((k+1)\ln e)^{(k+1)\ln 2}}, $$

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