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MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
did you try the ratio test?
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
\[L=\lim_{n\to\infty}\lefta_{n+1}\over a_n\right=\lim_{n\to\infty}\sqrt{2}>1\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
use induction to show that \[a_n\le 2\] for all \(n\)
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
So I just calculate its lowest upper bound limit by plugging in n?
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
oh nvm it cant be greater than 2 because 2= root of 4, right?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{2}\le 2\] assume \(a_n\le 2\) \(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
then use monotone convergence theorem
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
"...a function is bounded above if its increasing" that is not true
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I mean sequence* I thought if it was a sequence that was only increasing it was bounded above
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[a_n=n\] is an increasing sequence
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
it is not bounded above
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
if the sequence is bounded and it is either increasing or decreasing then the sequence converges.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
thank you for your patience!
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
Monotone convergence theorem
 one year ago
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