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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 $+\infty$

  • 6
    Show that for instance $a_i\geq 2^{i-1}$, since the powers of two diverge, you must have that the sequence diverges.2012-04-24
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    Have you written the terms of the sequence out explicitly?2012-04-24
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    As @David is suggesting, you have that $a_1=1$ $a_2=2$ $a_3=4$ $a_4=16$2012-04-24
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    I suppose that the succession goes to infinity with some terms that I found2012-04-24
  • 0
    Since every term is $\ge 1$, it diverges...2012-04-24
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    And what criteria I use to mean that diverges to +∞2012-04-24
  • 1
    @Emil, this is a sequence, not a sum.2012-04-24

1 Answers 1

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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

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