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Can anybody help me with this sequence question?

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I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]
did you try the ratio test?
\[L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|=\lim_{n\to\infty}\sqrt{2}>1\]

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Oops my post erased. My book / cal class does not and will not go over the ratio test I believe. I found the equation myself ( i mean the formula), when I use \[\lim_{n \rightarrow \infty} f(a _{n} ) = f(l)\], the answer it gives me is right but induction doesn't seem to work because I don't know if its bounded above ( well I'm assuming its above because it's increasing)
use induction to show that \[a_n\le 2\] for all \(n\)
So I just calculate its lowest upper bound limit by plugging in n?
oh nvm it cant be greater than 2 because 2= root of 4, right?
\[\sqrt{2}\le 2\] assume \(a_n\le 2\) \(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)
then use monotone convergence theorem
makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.
"...a function is bounded above if its increasing" that is not true
I mean sequence* I thought if it was a sequence that was only increasing it was bounded above
\[a_n=n\] is an increasing sequence
it is not bounded above
So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff
if the sequence is bounded and it is either increasing or decreasing then the sequence converges.
thank you for your patience!
Monotone convergence theorem

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