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anonymous
 4 years ago
how to prove the convergence of this sequence using the Bounded Monotone convergent theorem :
a_1 = sqrt(2) ;
a_(n+1) = sqrt(2*a_n) ;
and what is its limit ? ...
thanks~
anonymous
 4 years ago
how to prove the convergence of this sequence using the Bounded Monotone convergent theorem : a_1 = sqrt(2) ; a_(n+1) = sqrt(2*a_n) ; and what is its limit ? ... thanks~

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0limit is the easy part you have \[\lim_{n\rightarrow \infty} a_n=\lim_{n\rightarrow \infty}\sqrt{2a_n}\] if you call that limit say x, then you see that \[x=\sqrt{2x}\]meaning \[x^2=2x\] or \[x=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it possible to prove the fact the sequence is strictly increasing by doing the ratio test ? \[a_n+1/a_n = \sqrt(2*a_n) /a_n = \sqrt(2/a_n) =....\] but the problem is i don't know how to continue from the ... part. ( to show that the original question is strictly increasing i need to show \[a_n+1/a_n > 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you need bounded you need induction. you can in fact assert that \[a_n<2\] for all n step 1 of the inductions is \[a_1=\sqrt{2}<2\] and then assume it is true for all k < n and then \[a_n=\sqrt{2a_{n1}}\leq \sqrt{2\times 2} \text{ by induction } <\sqrt{4}=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0monotone i think again by induction, this time with \[a_{n+1}a_{n}=\sqrt{2a_n}\sqrt{2a_{n1}}\] then use \[\sqrt{a}\sqrt{b}=\frac{ab}{\sqrt{a}+\sqrt{b}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you get \[\frac{2(a_na_{n1})}{\sqrt{2a_n}+\sqrt{2a_{n1}}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0induction will show that this is greater than zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm i see, yeah so this does prove a_n is increasing XD~! thank you sooooo much XD~!
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