How can we use sequence $x_{n+1}=3-\frac{1}{x_n}$ , when $x_1=1 $ to evaluate $\sqrt5$?

Question

How can we use the sequence $x_{n+1}=3-\frac{1}{x_n}$ , when $x_1=1, $ to evaluate $\sqrt5$ ?

What I tried:

$x_1=1$

$x_2=2$

$x_3=2.5$

by excel spreadsheet I was able to get the converging value as $2.618033989$

How can I use this to evaluate $\sqrt5$,

Highly appreciated if someone can give me a hint to work this out.

Answer 1

Step 1

Show that $3\geq x_n\geq 1$ for all $n=1,2,\cdots.$

Notice $3\geq x_1\geq 1$. Assume $3 \geq x_k\geq 1.$ Then $3 \geq 3-\dfrac{1}{3}\geq x_{k+1} \geq 2 \geq 1.$ By induction, done.

Step 2

Show that $x_{n+1}\geq x_n$ for $n=1,2,\cdots.$

Notice $x_2=2 \geq x_1.$ Assume $x_{k+1} \geq x_{k}.$ Then $x_{k+2}=3-\dfrac{1}{x_{k+1}}\geq 3-\dfrac{1}{x_k}=x_{k+1}$. By induction, done.

Step 3

Show that $x_n$ is convergent.

Since $x_n$ is increasing and bounded above, hence $x_n$ is convergent.

Denote $\lim\limits_{n \to \infty}x_n=L.$ Then take the limits of the recursion formula. We have $L=3-\dfrac{1}{L},$ namely $L^2-3L+1=0.$ Thus, $L=\dfrac{3\pm \sqrt{5}}{2}.$ But since $x_n \geq 1,$ then $L \geq 1$. Hence $L=\dfrac{3+\sqrt{5}}{2}.$

Answer 2

When it converges you have that $x_{n+1}=x_n$ so we would have

$x=3-\frac{1}{x}$ which is the quadratic

$x^2-3x+1$

solving the quadratic gives us

$x=\frac{3 \pm \sqrt{3^2-4(1)(1)}}{2}$

or

$x=\frac{3 \pm \sqrt5}{2}$

the positive sign fits with what you found

$x=\frac{3 + \sqrt5}{2}=2.618033989$

so solving this gives you

$\sqrt{5}=2.236...$

Answer 3

This sequence tends to some $l$ such that$$l=3-\dfrac{1}{l}\\l^2-3l+1=0\\l=\dfrac{3\pm\sqrt{5}}{2}$$where $\dfrac{3-\sqrt{5}}{2}$ is invalid. So the sequence converges to $$l=\dfrac{3+\sqrt{5}}{2}$$and you can use this to attain some approximation of $\dfrac{3+\sqrt{5}}{2}$ hence $\sqrt 5$