# Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent [duplicate]

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 .

• Consider Cauchy's condensation test Mar 2 at 8:21

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.
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.
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.
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.
• How do you prove $(\ln \ln n)^2 \leq 4e^{-2} \ln n$ ? Mar 2 at 10:36
• @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 $4e^{-2}$.