Question #302080

let X1=1 and Xn+1=sqrt(3xn) for all n€N show that (Xn) is bounded monotone and converges to 3

1
Expert's answer
2022-02-28T11:43:11-0500

ANSWER.

Using the method of mathematical induction , we will show that xn<3x_{n}<3 for every nn .

Observe, that x1=1<3.x_{1}=1<3. From the inductive assumption xk<3x_{k}<3 it follows that xk+1=3xk<33=3x_{k+1}=\sqrt{3x_{k}}<\sqrt{3\cdot3 }=3 . So xn<3x_{n}<3 for every nn by the mathematical induction.

Since xn>0x_{n}>0 and xn3<0x_{n}-3<0 ,then xn(xn3)<0x_{n}\cdot (x_{n}-3)<0 or 3xn>xn23x_{n}>x_{n}^{2} . Hence,

xn+1=3xn>xnx_{n+1}=\sqrt{3x_{n}}>x_{n} .So the sequence is increasing. Since {xn}\left \{ x_{n} \right \} is increasing and bounded above limnxn=a\lim _{n\rightarrow\infty}x_{n}=a exists. The function f(x)=3xf(x)=\sqrt{3x} is continuous, therefore

a=limnxn+1=limn3xn=3limnxn=3aa=\lim_{n\rightarrow\infty}x_{n+1}=\lim_{n\rightarrow\infty}\sqrt{3x_{n }}=\sqrt{3\cdot \lim_{n\rightarrow\infty}x_{n }}=\sqrt{3a} . Thus a2=3aa^{2}=3a .

Since xn>x1=1x_{n}>x_{1}=1 for every nn we must have a1a\ge1 . Therefore a=3a=3 and limnxn=3\lim_{n\rightarrow\infty}x_{n}=3 .


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