anonymous
  • anonymous
Can anybody help me with this sequence question?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]
anonymous
  • anonymous
did you try the ratio test?
anonymous
  • anonymous
\[L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|=\lim_{n\to\infty}\sqrt{2}>1\]

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anonymous
  • anonymous
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)
Zarkon
  • Zarkon
use induction to show that \[a_n\le 2\] for all \(n\)
anonymous
  • anonymous
So I just calculate its lowest upper bound limit by plugging in n?
anonymous
  • anonymous
oh nvm it cant be greater than 2 because 2= root of 4, right?
Zarkon
  • Zarkon
\[\sqrt{2}\le 2\] assume \(a_n\le 2\) \(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)
Zarkon
  • Zarkon
then use monotone convergence theorem
anonymous
  • anonymous
makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.
Zarkon
  • Zarkon
"...a function is bounded above if its increasing" that is not true
anonymous
  • anonymous
I mean sequence* I thought if it was a sequence that was only increasing it was bounded above
Zarkon
  • Zarkon
\[a_n=n\] is an increasing sequence
Zarkon
  • Zarkon
it is not bounded above
anonymous
  • anonymous
So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff
Zarkon
  • Zarkon
if the sequence is bounded and it is either increasing or decreasing then the sequence converges.
anonymous
  • anonymous
thank you for your patience!
Zarkon
  • Zarkon
Monotone convergence theorem

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