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Let $\{a_{n}\}$ a sequence such that $a_{n +1}=2^{a_{n}}$, $a_{1}=1$ show that $\{a_{n}\}$ diverges to $+\infty$

hint:

It would have to prove by induction that: $a_{n}\geq 2^{n-1}$, $n = 2,3, ...$

Using the inequality $2^{n-1}=(1 +1)^{n-1}=1+(n-1)+\cdots\geq n$ (if $n\geq 2$)

Could they please explain this exercise?

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Explain the exercise, explain the hint, explain how to use the hint? Which? –  Arturo Magidin May 3 '12 at 5:13

1 Answer 1

up vote 2 down vote accepted

For any $n \geq 1$, $$a_{n+1}=2^{a_n} \geq a_n.$$ Hence $\{a_n\}_n$ is steadily increasing, and $L=\lim_{n \to +\infty} a_n$ exists, finite of infinite. If $L<+\infty$, then $L=2^L$, and this is impossible: sketch the graphs of $L \mapsto 2^L$ and remark that it lies above $L \mapsto L$. Hence $L=+\infty$.

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