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MarcLeclair
 one year ago
Can anybody help me with this sequence question?
MarcLeclair
 one year ago
Can anybody help me with this sequence question?

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MarcLeclair
 one year 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 \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]

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

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

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

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

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

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

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

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

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

MarcLeclair
 one year 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
 one year 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.

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

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