Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent 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 .
 A: 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.
A: 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.
A: 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.
A: 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.
