TITLE: The limit of $a_1^{a_2^{.^{.^{.^{a_n}}}}}-a_n^{a_{n-1}^{.^{.^{.^{a_1}}}}}$ as $n\to\infty$.
QUESTION [6 upvotes]: If this is a duplicate in any way, I'm sorry.
The Question:

For what $(a_i)_{i\in \Bbb N}\in \Bbb R_+^{\Bbb N}$ does
$$L:=\lim_{n\to \infty}a_1^{a_2^{.^{.^{.^{a_n}}}}}-a_n^{a_{n-1}^{.^{.^{.^{a_1}}}}}$$
exist and what values does $L$ take?

Thoughts:
Of course, $L$ exists and would be $0$ when $a_i=1$ for all $i\in\Bbb N$. Also $L=1$ if
$$a_i=\begin{cases}
2 &: i=1 \\
1 &: i\neq 1.
\end{cases}$$

REPLY [2 votes]: This is a partial answer, it address what value of $L$ can take.
It turns out $L$ can take any real number as value.
For simplicity of typesetting, I will use Knuth's up-arrow notation to represent any tower of exponentiation. Furthermore, we will assume the up-arrows are right-associative. More precisely,
$$a_1 \uparrow a_2 \uparrow \cdots \uparrow a_n
\;=\; a_1 \uparrow \left( a_2 \uparrow \left(\cdots \uparrow a_n\right)\right)
\;\stackrel{def}{=}\;a_1^{a_2^{.^{.^{.^{a_n}}}}}
$$
For any $z \in \mathbb{R}$, rewrite $z$ as $x-y$ for some $x, y > 0$. Let $(y_k)_{k \ge 0}$ be any sequence of positive numbers such that $y_0 = 1$ and $y_n \to y$ as $n \to \infty$.
Consider the sequence $(a_k)_{k > 0}$ defined by
$$a_1 = x,\quad a_2 = 1\quad\text{and}\quad a_{n+2} = y_{n}^{1/y_{n-1}}\quad\text{for}\quad n > 0 $$
It is easy to see for any $n > 0$,
$$a_1 \uparrow a_2 \uparrow \cdots \uparrow a_{n+2}
= x \uparrow ( 1 \uparrow ( \cdots ) ) = x \uparrow 1  = x$$
Furthermore,
$$\begin{align}
  a_{n+2} \uparrow \cdots \uparrow a_5 \uparrow a_4 \uparrow a_3 \uparrow a_2 \uparrow a_1
&=  y_n^{1/y_{n-1}} \uparrow \cdots \uparrow y_3^{1/y_2} \uparrow y_2^{1/y_1}\uparrow y_1^{1/y_0} \uparrow 1 \uparrow x\\
&=  y_n^{1/y_{n-1}} \uparrow \cdots \uparrow y_3^{1/y_2} \uparrow y_2^{1/y_1}\uparrow y_1 \uparrow 1 \\
&=  y_n^{1/y_{n-1}} \uparrow \cdots \uparrow y_3^{1/y_2} \uparrow y_2^{1/y_1}\uparrow y_1 \\
&= y_n^{1/y_{n-1}} \uparrow \cdots \uparrow y_3^{1/y_2} \uparrow y_2 \\
&\;\;\vdots\\
&= y_n 
\end{align}
$$
For this particular choice of $a_k$, we have
$$L = \lim_{n\to\infty} 
\left( 
a_1 \uparrow a_2 \uparrow \cdots \uparrow a_{n+2} -
a_{n+2} \uparrow a_{n+1} \uparrow \cdots \uparrow a_1
\right)
 = \lim_{n\to\infty} (x - y_n) = x - y = z$$
From this, we can conclude $L$ can take any real number as value.