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Possible Duplicate:
$\{a_{n}\}$ diverges to $+\infty$

Let $ a_ {n} $ a sequence such that $ a_ {n + 1} = 2 ^ {a_ {n}} $, $ a_ {1} = 1 $. Show that $ a_ {n} $ diverges at $ \infty $

My attempt:

Show that is growing by induction 1) $ a_{1}, a_{2}=2$, $a_{1}

2) We assume that $ a_ {n-1} . $ \Rightarrow a_ {n} then the sequence is increasing -Prove that is not bounded. Let $ M = 2 ^ {n-1} $, to ​​show that $ S_{n} \geq M \forall n \geq N, N \in \aleph $. By induction: -For $ n = 1 $, $ a_ {2} = 2 \geq 2 ^ {a_ {1}} = $ 2 -We assume that is true for n $ a_ {n} \geq 2 ^ {n-1} $ On the other hand we have to apply Newton's binomial $ 2 ^ {n-1} = (1+1) ^ {n-1} = 1 (n-1) ... (n-1) 1 = n ... n \geq n $ $ \Rightarrow 2 ^ {n-1} \geq n $. Then $ a_ {n} \geq 2 ^ {n-1} \geq n \Rightarrow a_ {n} \geq n $ Hence $ 2 ^ {a_ {n}} \geq 2 ^ {n} $ then the sequence is not bounded. Therefore the sequence diverges to $ \infty $

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    I don't really understand your notation here. Should that be $a_{n+1}$ in the definition of the sequence? The first step of your argument does not quite follow (unless I misread it). I would prove by induction that the sequence is increasing. If you suppose by way of contradiction that the sequence is bounded, then you know that it in fact converges since it is monotone. Call the limit $x$; it would have to satisfy $ x = 2^x$ by taking limits on both sides of the defining equation for the sequence. No real number satisfies that equality. The approach you gave works just as well though.2012-05-04
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    There are some errors in the second part, but it's hard to tell how serious they are because of the very sloppy presentation. At the very least the two equations after $(1+1)^{n-1}$ are wrong.2012-05-04
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    To show it is unbounded, it suffices to show that for every positive integer $M$, there is an integer $n$ such that $a_n > M$. But this is easy, $n = M+1$ will work. An induction proof shows that $a_n \geq n$ since $a_{k+1} = 2^{a_k} \geq 2^k = (1+1)^k \geq 1+ (k)(1) = k+1$ (using the induction hypothesis on $k$, the base case is obvious).2012-05-04

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