Determine whether $\sum_{k=2}^\infty\frac1{(\ln k)^{\ln k}}$ is convergent or divergent. I have the series
$$\sum_{k=2}^\infty\frac1{(\ln k)^{\ln k}}$$
and am trying to find whether it is convergent or divergent. Along with the question, I received a hint stating that
$$\ln k=e^{\ln\ln k}$$
I simplified to make
$$\sum_{k=2}^\infty \frac1{e^{\ln\ln k\ln k}}$$
given the hint. I am unsure of what series test to use in order to proceed, due to the unique nature of the question.
 A: Since $e^{\ln k}=k$ for $k>0$:
$$\sum_{k=2}^\infty\frac1{e^{\ln\ln k\ln k}}=\sum_{k=2}^\infty k^{-\ln\ln k}$$
$\ln\ln k$ is a strictly increasing function for $k>1$, so there exists an integer $n$ with $\ln\ln m>2$ for all $m\ge n$. Thus, the series splits into a finite head and a tail that can be shown convergent by the limit comparison test:
$$\sum_{k=2}^\infty k^{-\ln\ln k}=\sum_{k=2}^{n-1} k^{-\ln\ln k}+\sum_{k=n}^\infty k^{-\ln\ln k}\le\sum_{k=2}^{n-1} k^{-\ln\ln k}+\sum_{k=n}^\infty k^{-2}<\sum_{k=2}^{n-1} k^{-\ln\ln k}+\frac{\pi^2}6<\infty$$
Therefore the original series converges.
A: The easiest is to use the condensation test, $\sum a_n$ converges iff $$\sum 2^na_{2^n}$$ converges. In this case you get 
$$\sum \frac{2^n}{n^n}$$ which converges, by root test or others.
A: Using Ermakoff's test:
$$\lim_\limits{x\to\infty} \frac{e^xf(e^x)}{f(x)}=\lim_\limits{x\to\infty} \frac{e^x\cdot \frac{1}{x^x}}{\frac{1}{(\ln x)^{\ln x}}}=\lim_\limits{x\to\infty} \frac{e^x\cdot (\ln x)^{\ln x}}{x^x}=0<1.$$
So it converges.
A: More generally,
generalizing
Parcly Taxel's proof,
if
$f(k)$
is positive,
strictly increasing,
and eventually $> e$,
then
$\sum_{k=2}^\infty\frac1{f(k)^{\ln k}}
=\sum_{k=2}^\infty\frac1{e^{\ln(f(k))\ln k}}
=\sum_{k=2}^\infty\frac1{k^{\ln(f(k))}}
$
and this converges
by comparison with
$\frac1{k^p}$
with
$\ln(f(k)) > 1$.
