Could someone help me through this problem?
Let $ a_ {n} $ a sequence such that $ a_ {n +1} = 2 ^ {a_ {n}} $, $ a_ {1} = 1$ show that $a_ {n}$ diverge to $+\infty$
Could someone help me through this problem?
Let $ a_ {n} $ a sequence such that $ a_ {n +1} = 2 ^ {a_ {n}} $, $ a_ {1} = 1$ show that $a_ {n}$ diverge to $+\infty$
The $a_n$ makes the tetration $a_n = \begin{cases} \underbrace{2^{2^{\cdot^{\cdot^{2}}}}}_{n-1}, & \mbox{if } a > 1 \\ 1, & \mbox{if } a = 1. \end{cases} = {^{n}2}$.
Euler proved that infinite tetrations in the form
$\lim_{n \rightarrow \infty} {^{n}x} = x^{x^{\cdot^{\cdot^{x}}}}$
only converges for $e^{−e} ≤ x ≤ e^{1/e}$.
Now, $2 > e^{1/e} \approx 1.44$, thus
$\lim_{n \rightarrow \infty} a_n = \infty$