Check convergence of $\sum^{\infty}_{n=1} \frac{1}{(\ln\ln n)^{\ln n}}$ Check convergence of 
$$\sum^{\infty}_{n=1}\frac{1}{(\ln \ln n)^{\ln n}}.$$
Please verify my solution below. 
 A: There is $n_0$ such that
$$\sum^{\infty}_{n=n_0}\frac{1}{(\ln \ln n)^{\ln n}}<\sum^{\infty}_{n=n_0}\frac{1}{3^{\ln n}}=\sum^{\infty}_{n=n_0}\frac{1}{n^{\ln 3}}$$
and this converges since $\ln 3>1$
Q.E.D.
A: $$\sum^{\infty}_{n=1}\frac{1}{(\ln\, \ln\, n)^{\ln\, n}}=\sum^{\infty}_{n=1}\frac{1}{\exp( \ln\, \ln\ln\, n\,*\ln\,, n)}=\sum^{\infty}_{n=1}\frac{1}{\exp(\ln\, n\,*\ln\, \ln\, \ln\, n)}=\sum^{\infty}_{n=1}\frac{1}{n^{\ln\, \ln\, \ln\, n}}$$
For a lagre n, $$\frac{1}{n^{\ln\, \ln\, \ln\, n}} <  \frac{1}{n^2}$$
So the series converge by comparison test
A: Use the inequality
$$
\frac{x-1}{x^p}<\ldots<\frac{x-1}{x^2}<\frac{x-1}{x}<\log x , \quad \forall \,x>1, \,\forall p\in\mathbb{N}\backslash\{1,0\}.
$$
and the fact that $\ln$ is an increasing function. 
We have  for $n$ large, 
\begin{align}
\left|\frac{1}{\left[\log\big(\log n\big)\right]^{(\log n)}}\right|\leq
&
\frac{1}{\left|\log\circ\log n\right|^{\big(\frac{n-1}{n^p}\big)}}
\\
\leq
&
\frac{1}{\left|\log\big(\frac{n-1}{n}\big)\right|^{\big(\frac{n-1}{n^p}\big)}}
\\
\leq
&
\frac{1}{\left|\frac{\big(\frac{n-1}{n}\big)-1}{\big(\frac{n-1}{n}\big)}\right|^{\big(\frac{n-1}{n^p}\big)}}
\\
=
&
\frac{1}{\left(\frac{n}{n-1}\right)^{\big(\frac{n-1}{n^p}\big)}}
\\
=
&
\bigg(1-\frac{1}{n}\bigg)^{\frac{n-1}{n^p}}
\\
=
&
\left[\bigg(1-\frac{1}{n}\bigg)^{n}\right]^{\frac{1}{n^p}}
\\
\end{align}
And by root test, if $a_n= \left[\bigg(1-\frac{1}{n}\bigg)^{n}\right]^{\frac{1}{n^p}}$ we have 
$
\lim_{n\to \infty}\sqrt[n]{a_n}<1 
$
for $p$ enogh large.
By comparation test  the serie 
$$
\sum_{n=1}^{\infty}\frac{1}{\left[\log\big(\log n\big)\right]^{(\log n)}}
$$
converge.
