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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- Sep 22nd 2009, 04:31 PMcgiulzShowing a sequence tends to infinity
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. - Sep 23rd 2009, 12:32 AMpaweld
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. - Sep 23rd 2009, 06:09 PMcgiulz
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 $\displaystyle n$?