TITLE: Power tower inequality
QUESTION [7 upvotes]: I want to prove the following power tower inequality:
$$
3 \uparrow \uparrow 100 > 4 \uparrow \uparrow 99
$$
but I don't know how to do this. I think that induction will not work, because I think there will be an $N$ for which
$$
3 \uparrow \uparrow N < 4 \uparrow \uparrow (N-1)
$$
Could anyone help me in the right direction? Please don't answer a full solution, but I do need a hint.

REPLY [2 votes]: Actually, $4 \uparrow \uparrow n < 3 \uparrow \uparrow (n+1)$ for all $n \ge 0$. 
In fact, it can be shown that if $x$ and $y$ are such that $1 < x < y$, then
$$y \uparrow \uparrow n \ \ < \ \ x \uparrow \uparrow (n+c)\ \ \quad(n \ge 0)$$
where $c$ is a postive integer depending only on $x$ and $y$. 
Here's a proof-outline (adapted from this posting and followups):

Suppose $1 < x < y$, and let $a_n = {\log_x}^n(y\uparrow\uparrow n)$ for $n \ge 0$. Consider the sequence $(a_n)$ as $n \to \infty.$
Show that $(a_n)$ is strictly increasing, using properties of the logarithm.
Show that $(a_n)$ has a limit, say $A$, using the Mean Value Theorem to prove extremely rapid convergence.
Note that $y\uparrow\uparrow n = (x\uparrow)^n a_n < (x\uparrow)^n A \le (x\uparrow)^n (x\uparrow)^c 1$ = $x\uparrow\uparrow (n+c)$, where $c$ is the least integer satisfying $A \le (x\uparrow)^c 1 = x\uparrow\uparrow c$. 

Here are some particular cases:
x   y   c   A
--  --  --  ----------------
2   3   2   2.44402146148920
2   4   2   3.17037617633756
2   5   2   3.68091002494335
2   6   3   4.07723742182623
e   3   1   1.22172930187025
3   4   1   1.51107202382304   <-- your case
3   5   1   1.85474212525557
4   5   1   1.28188454071981