2
$\begingroup$

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 $

  • 0
    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

0 Answers 0