Define a sequence recursively by $\displaystyle a_{n+1} = 2a_{n}^2$.
Given $\displaystyle a_{0} > 1/2$, show lim $\displaystyle a_{n} = \infty$.
Note: It can easily be shown that $\displaystyle a_{n} > 1/2$ so just assume this is given.
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Define a sequence recursively by $\displaystyle a_{n+1} = 2a_{n}^2$.
Given $\displaystyle a_{0} > 1/2$, show lim $\displaystyle a_{n} = \infty$.
Note: It can easily be shown that $\displaystyle a_{n} > 1/2$ so just assume this is given.
Let $\displaystyle b_n=2 a_n $.
We have the following reccurence for sequence $\displaystyle b_n$:
$\displaystyle
b_{n+1}=b_n^2
$
But for this sequence explicit formula can be easily obtained:
$\displaystyle
b_n=b_0^{2^n}
$
Because $\displaystyle b_0 > 1$ the sequence is not convergent.
Hey, sorry for the late response, thanks though.
So I can just assume $\displaystyle b_{n} > 1/2$ and just say since,
$\displaystyle b_{n} < a_{n}$ for all $\displaystyle n$?
