let X1=1 and Xn+1=sqrt(3xn) for all n€N show that (Xn) is bounded monotone and converges to 3
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Expert's answer
2022-02-28T11:43:11-0500
ANSWER.
Using the method of mathematical induction , we will show that xn<3 for every n .
Observe, that x1=1<3. From the inductive assumption xk<3 it follows that xk+1=3xk<3⋅3=3 . So xn<3 for every n by the mathematical induction.
Since xn>0 and xn−3<0 ,then xn⋅(xn−3)<0 or 3xn>xn2 . Hence,
xn+1=3xn>xn .So the sequence is increasing. Since {xn} is increasing and bounded above limn→∞xn=a exists. The function f(x)=3x is continuous, therefore
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