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

is divergent . What inequality can we use here. i tried various method but none of these give any result. Any hint please .

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    $\begingroup$ Consider Cauchy's condensation test $\endgroup$
    – TravorLZH
    Mar 2 at 8:21
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By the integral test: $$ \sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}\sim \int_2^\infty\frac{dx}{(\log x)^{\log(\log x)}}\stackrel{\log x\mapsto u}= \int_{\log2}^\infty\frac{e^udu}{u^{\log(u)}}\stackrel{\log u\mapsto t}= \int_{\log(\log2)}^\infty e^{e^t-t^2+1}d t. $$ which clearly diverges.

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Take the exponential of the denominator. It can be seen that $\log\log(n) n$ is smaller than $e^n$ for large enough $n$. Since $x \mapsto e^x$ is monotonically increasing, this means the deonminator is smaller than $n$. The series can then be compared with the harmonic series.

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By Cauchy condensation test, the given series converges if and only if $\sum_{n=1}^\infty a_n$ does, where $$a_n=\frac{2^n}{(n\log 2)^{\log(n\log 2)}}.$$ But $a_n\not\to 0$ (and in fact $a_n\to\infty$) as $n\to\infty$. Hence, the given series diverges.

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Note that for all $n\geq 2$, $$ (\log n)^{\log \log n} = e^{(\log \log n)^2 } \le e^{4e^{ - 2} \log n} < e^{\log n} = n. $$ Consequently $$ \sum\limits_{n = 2}^\infty {\frac{1}{{(\log n)^{\log \log n} }}} > \sum\limits_{n = 2}^\infty {\frac{1}{n}} , $$ and we are done since the harmonic series is divergent.

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  • $\begingroup$ How do you prove $(\ln \ln n)^2 \leq 4e^{-2} \ln n$ ? $\endgroup$ Mar 2 at 10:36
  • $\begingroup$ @GabrielRomon You consider $(\log\log x)^2/\log x$ for $x\geq 2$. By differentiation, it is not hard to show that it has a maximum at $x=e^{e^2}$. The value of this maximum is $4e^{-2}$. $\endgroup$
    – Gary
    Mar 2 at 11:02

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