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anonymous
 3 years ago
Can anybody help me with this sequence question?
anonymous
 3 years ago
Can anybody help me with this sequence question?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you try the ratio test?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[L=\lim_{n\to\infty}\lefta_{n+1}\over a_n\right=\lim_{n\to\infty}\sqrt{2}>1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops 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)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So I just calculate its lowest upper bound limit by plugging in n?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh nvm it cant be greater than 2 because 2= root of 4, right?

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

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1then use monotone convergence theorem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1"...a function is bounded above if its increasing" that is not true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean sequence* I thought if it was a sequence that was only increasing it was bounded above

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1\[a_n=n\] is an increasing sequence

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1if the sequence is bounded and it is either increasing or decreasing then the sequence converges.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you for your patience!

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1Monotone convergence theorem
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