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Is the series

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

convergent?

I know that $ \log(\log n) >2$. Now $\log(\log(\log (n))) > \log 2$. After that I am not able to proceed further.

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Note that since $x\ge \log^2(x)$ for $x\ge \log(2)$, we have

$$\begin{align} \int_2^L \frac{1}{(\log(x))^{\log(\log(x))}}\,dx&\overbrace{=}^{x\mapsto e^x}\int_{\log(2)}^{\log(L)}\frac{e^x}{x^{\log(x)}}\,dx\\\\ &=\int_{\log(2)}^{\log(L)}e^{x-\log^2(x)}\,dx\\\\ &\ge \log(L/2) \end{align}$$

Hence, the integral test guarantees that the series of interest diverges.

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  • $\begingroup$ I can't manage series by integrals abd I can't follow but it seems a nice proof! $\endgroup$ – gimusi Apr 9 '18 at 19:34
  • $\begingroup$ @gimusi See This article on the integral test. It is effectively the same as the Cauchy Condensation test. And a simple proof of the latter is based on the former. $\endgroup$ – Mark Viola Apr 9 '18 at 19:36
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Apr 10 '18 at 12:37
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Hint: $a^b$ $= (e^{\log a})^b$ $= e^{b\log a}$. Applying this to your denominator turns it into $e^{(\log\log n)^2}$, and now you should be able to compare with $e^{\log n}$...

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  • $\begingroup$ (+1) I don't know why this hasn't received many more up votes. $\endgroup$ – Mark Viola Apr 9 '18 at 20:00
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HINT

Let use the Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

that is

$$\sum_{n=2}^\infty \frac{1}{(\log n)^{\log\log n}} \ge \frac12\sum_{n=2}^\infty \frac{2^n}{(\log 2^n)^{\log\log 2^n}}=\frac12\sum_{n=2}^\infty \frac{2^n}{(n\log 2)^{\log (n\log 2)}}$$

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  • $\begingroup$ im not getting @gimusi...is there is any simple method is there...? $\endgroup$ – user396850 Apr 9 '18 at 19:12
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    $\begingroup$ @Stupid that's not so bad! $\endgroup$ – gimusi Apr 9 '18 at 19:13
  • $\begingroup$ Why dont you finish the question ? $\endgroup$ – Rene Schipperus Apr 9 '18 at 19:17
  • $\begingroup$ @ReneSchipperus it's just an hint and the last series seems seriously aimed to diverge $\endgroup$ – gimusi Apr 9 '18 at 19:20
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    $\begingroup$ @Stupid To finish note that $$\frac{2^n}{(n\log 2)^{\log (n\log 2)}}\ge\frac{2^n}{n^{\log (n)}}$$ $\endgroup$ – gimusi Apr 9 '18 at 19:31

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