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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 $+\infty$

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Show that for instance $a_i\geq 2^{i-1}$, since the powers of two diverge, you must have that the sequence diverges. – Daniel Montealegre Apr 24 '12 at 18:26
Have you written the terms of the sequence out explicitly? – David Mitra Apr 24 '12 at 18:27
As @David is suggesting, you have that $a_1=1$ $a_2=2$ $a_3=4$ $a_4=16$ – Pedro Tamaroff Apr 24 '12 at 18:29
I suppose that the succession goes to infinity with some terms that I found – Chalie Her Apr 24 '12 at 18:31
@Emil, this is a sequence, not a sum. – Neal Apr 24 '12 at 22:34
up vote 2 down vote accepted

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

$$\lim_{n \rightarrow \infty} a_n = \infty$$

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