Consider for $x>e^e$ and $0<A<5$:
$$g_A(x) = \exp(\ln(x) \ln(\ln(x))^A)$$
I want to find a function $f_A(x)$ ,
$$f_A(x) = a_0(A) + a_1(A) x + a_2(A) x^2 + ... = \sum_{n=0}^{\infty} a_n(A) x^n$$
Where for every $n>0$ :
$$0 < a_{n+1}(A) < a_n(A)$$
and those $a_i$ are constants only depending on $A$, not on $x$.
And also
$$\lim_{x \to +\infty} \frac{g_A(x)}{f_A(x)} = 1$$
So basically $f_A(x)$ is a very good asymptotic for $g_A(x)$ and we want the Maclauren coefficients of $f_A(x)$ under the given restrictions.
Ok maybe that is asking alot.
Let us consider the simpler case and degeneralize,
Lets say $A=2$ then
$$g(x) = \exp(\ln(x) \ln(\ln(x))^2)$$
$$f(x) = a_0 + a_1 x + a_2 x^2 + ... = \sum_{n=0}^{\infty} a_n x^n$$
Where for every $n>0$ :
$$0 < a_{n+1} < a_n$$
And
$$\lim_{x \to +\infty} \frac{g(x)}{f(x)} = 1$$
What are good approximations of $a_n$ ?
I know
$$h(x) = \sum_{n=0}^{\infty} h_n x^n $$
with $h_n = c_0 c_1^{- n^{c_2}}$ is close to $\exp(c_3 \ln(x)^{c_4})$ for the appropriate positive constants $c_0,c_1,c_2,c_3,c_4$
Therefore $a_n < c_0 c_1^{- n^{c_2}}$ is for certain.
So some guesses are
$$a_n = \exp(- \exp(n)) ?$$
$$a_n = \exp(- n!)$$
$$a_n = \exp(- \exp(\ln(n)^2)) ?$$
How to solve this ?
What I tried.
- Integral analogue
$$g(x) = \int_1^{+\infty} a(t) x^t dt$$
Contour integral ideas
The heuristic (known as " fake function theory " by some)
$$a_n < Min [ \frac{g*(x)}{x^n} ]$$
Where $g*$ is the same as $g$ but $\ln(x)$ is replaced by an asymptotic for it that is defined well for all $x>0$ such as $ArcSinh(x/2)$.
Maybe I made a mistake but no sharp results came to me from those.
edit
I think
$$O(\exp(\exp(\ln(x) \ln(\ln(x))^2))) = \sum_{n=0}^{\infty} k_n x^n$$ and $$0 < k_n$$
holds for
$$\frac{1}{k_n} = O((n \ln(\ln(n))^l)^{m n \ln(\ln(n))^l})$$
and some real $l>1/4$ and real $m>0$.
That looks similar. Maybe it could help. Assuming it is true ofcourse. Even so that is not the most efficient way probably. Just sharing the idea.
Not much of an answer but based on the heurstic 3 " fake function theory " (described in the OP) , there is the asymptotic formula :
For the conditions (more or less, the theory is unfinished)
$$ F(x) < \sqrt x $$ and $F(x)$ strictly increasing and $F'(x)$ strictly decreasing ;
$$ x^{F(x)} = O( \sum \frac{x^n}{G(n-1)} )$$
where $G(x)$ is the functional inverse of $F(x)$ and $O(*)$ is big-O notation.
This gives as estimate here
$$g_A(x) = x^{\ln(\ln(x))^A} = O( \sum \frac{x^n}{\exp(\exp((n-1)^{1/A}))} ) $$
I think I can push it a bit further, but this estimate is brutal.
A much better estimate is desired.
See also this topic :
https://mathoverflow.net/questions/219428/asymptotics-to-taylor-expansions
or the same one here at MSE :
Proof that $\oint_r d(x,N + n) < 0 $?
which implies when true ! for $x>>1$:
$$ \sum \frac{ x^n}{\sqrt n \ln(e+n) \exp(\exp((n-1)^{1/A}))} ) < x^{\ln(\ln(x))^A} < \sum \frac{\sqrt n \ln(e+n) x^n}{\exp(\exp((n-1)^{1/A}))} ) $$
But I do not want to force a specific viewpoint to the idea.
Maybe this is overkill and too complicated for the considered question and there exist simpler more efficient ways.
Just my 2 cents.
Btw $Min$ (minimum) means $Inf$ (infimum) and we usually compute it by taking the derivate of it and matching it to zero.
See also this link maybe to clarify :
Post nr 9 of " fake function theory " :
https://tetrationforum.org/showthread.php?tid=863&pid=6995#pid6995
( do not be confused by the semi-exp here ; f(f(x)) = exp(x) ... yeah it is at a teration forum , but the idea is very general so it works for general cases )