I have this series that I can't understand how to find it's character.
$$\tag{1} \sum_{n=4}^\infty \left(\frac{1}{\log(\log(n))}\right)^{\log(n)} $$To exercise, I solved a similiar series to get the ground how to solve the previous one.
$$\tag{2} \sum_{n=2}^\infty \left(\frac{1}{\log(n)}\right)^{\log(n)} $$
This is my solution:
$ a_n $ is a series made by positive terms, so it can't be unsolvable (diverges positively or converges).
$$ \log(n)^{\log(n)} = (e^{\log(n)})^{\log(\log(n))} = n^{\log(\log(n))} \ge n^{\log(3)} $$
$ \forall n >e^3$, $\log(3)>3$, $\log(\log(n))>\log(3)>1$. So we have that $$ \frac{1}{\log(n)^{\log(n)}} \le \frac{1}{n^{\log(3)}} $$ $\forall n> e^3 $.
And by the critery of asymptotic comparison by the fact that
$$\tag{3} \sum \frac{1}{n^{\log(3)}}$$ is the generalized harmonic series with $p=\log(3)>1 $ so it converge. $(3)$ converges so then $(2)$ converges.
I don't know how to use this to solve $(1)$, I can't find any inequality to get it right.
Best Answer
Your argument works more generally.
Given any positive function $f(n):$
$$f(n)^{\log n}=e^{\log f(n)\log n}=n^{\log f(n)}$$
Now, if $f(n)\to\infty$ as $n\to\infty$ we can get that $$\sum \left(\frac1{f(n)}\right)^{\log n}$$ converges, by your reasoning.
Even weaker still, if there is some $c>e$ and some $N,$ such that $f(n)\geq c$ for all $n\geq N,$ then the series converges.
So you only need $f(n)=\log\log n\geq 3$ for $n$ sufficiently large, say $n>e^{e^3}.$
If $f$ is non-decreasing, the series diverges if and only if $f(n)\leq e$ for all $n.$