Answer to Question #302080 in Calculus for sakshi

Question

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

Answer 1

ANSWER.

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

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

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

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

$a=\lim_{n\rightarrow\infty}x_{n+1}=\lim_{n\rightarrow\infty}\sqrt{3x_{n }}=\sqrt{3\cdot \lim_{n\rightarrow\infty}x_{n }}=\sqrt{3a}$ . Thus $a^{2}=3a$ .

Since $x_{n}>x_{1}=1$ for every $n$ we must have $a\ge1$ . Therefore $a=3$ and $\lim_{n\rightarrow\infty}x_{n}=3$ .