The limit of $a_1^{a_2^{.^{.^{.^{a_n}}}}}-a_n^{a_{n-1}^{.^{.^{.^{a_1}}}}}$ as $n\to\infty$.

Question

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 $$a_i=\begin{cases} 2 &: i=1 \\ 1 &: i\neq 1. \end{cases}$$

I'm tempted to say that $L=0$ when $a_i=a_1$ for all $i\in \Bbb N$. — Shaun, Jul 21 '17 at 08:00
I think $L$ would be $0$ as well if all $a_i$ are equal, right? — G. Ünther, Jul 21 '17 at 08:00
With complex exponents and complex bases, there is always the question of multivalues. How do you pick them? — Arthur, Jul 21 '17 at 08:06
I don't know if the limit is defined if two consecutive $a_i$ are zero. Because, what is $0^0$? — G. Ünther, Jul 21 '17 at 08:06
@Shaun May be restrict $a_i$ to be positive real number for each $i$ ? — pks, Jul 21 '17 at 08:35
Since $(a_i) = (x,1,1,\ldots) \leadsto L = x - 1$, so all $L > -1$ is possible. — achille hui, Jul 21 '17 at 08:56
Of similar flavor: The least value of $b$ such that $2^{3^{\cdots^a}}\leq b^{(b-1)^{\cdots^2}}$ — Bart Michels, Jul 21 '17 at 09:24
What would be interesting is if both of the power towers tended towards infinity but their difference did not, excluding the trivial constant sequence case. — Bob Krueger, Jul 21 '17 at 13:58

score 2 · Answer 1 · answered Jul 22 '17 at 13:25

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.

The limit of $a_1^{a_2^{.^{.^{.^{a_n}}}}}-a_n^{a_{n-1}^{.^{.^{.^{a_1}}}}}$ as $n\to\infty$.

The Question:

Thoughts:

1 Answers1

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