Upper bound of $x\log(x\log(x\log(x\log(...))))$ What is an upper bound (maybe in terms of $x$) of the following sequence as $n\to\infty$?
$$a_0 = x,$$
$$a_{n+1} = x \log a_n$$
 A: ** some hints**
Put $$f(t)=t\ln(t)$$
$$f'(t)=\ln(t)+1$$
If $0<x\le 1$ then
$a_1\le 0$ and $a_2$ is not defined.
If $1<x<e$ the sequence decreases until becoming undefinite. Some $a_k\le 1$.
If $x=e$, the sequence is stationary.
$$a_n=e$$
Finally, if $x>e$, the sequence will increase to infinity :
By induction, we show that
$$(\forall n>0) \;\; a_n>x$$
and by MVT
$$a_{n+2}-a_{n+1}=(a_{n+1}-a_n)f'(c)$$
$$=(a_{n+1}-a_n)(\ln(c)+1)>(a_{n+1}-a_n)(\ln(x)+1)$$
with $\ln(x)+1>2$.
thus
$$a_{n+1}-a_n>2^n(a_1-x)$$
A: The fixed-point iteration function $g(a)=x\ln(a)$ has derivative $$g'(a)=\frac{x}a.$$ Thus $g$ is contractive in the sense $0<g'(a)<1$ for $a>x$. 
The limit, if it exists, can be computed via $$a_*=e^b=bx\iff -be^{-b}=-\frac1x$$ which has solutions using the Lambert-W or poly-logarithmic functions $b=-W_0(-x^{-1})\in(0,1)$ and $b=-W_{-1}(-x^{-1})\ge-1$ for $x\ge e$. Then
$$
a_*=-xW_{-1}(-x^{-1})\ge x
$$
is the only fixed point inside the region $a\in [x,\infty)$ of contraction in the sense $|g(a)-a_*|<|a-a_*|$, and thus the fixed point of the iteration. All sequences $a_{k+1}=g(a_k)$ starting with $a_0\ge x$ will converge toward that point $a_*$.
A: For $t>0,m>1$, $$\log t\le m(t^{1/m}-1)\le mt^{1/m}.$$
So if $a_n>1$,
$$a_{n+1}=x\log a_n\le mxa_n^{1/m}$$
or
$$\log a_{n+1}\le \log m+\log x+\frac 1m\log a_n.$$
The homogeneous part gives
$$\log a_n= \frac{\log a_0}{m^n},$$ and a general solution is
$$\log a_n\le c\frac{\log a_0}{m^n}+\frac{m}{m-1}(\log m+\log x)$$
or
$$a_n\le \sqrt[m^n]{a_0^c}+\sqrt[m-1]{(\log m+\log x)^m}.$$
Not quite rigorous, but not quite useless...
