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 $
real-analysis
sequences-and-series