Chapter 8.1, Problem 56E

To determine

To show: that the sequence defined by $a_1=2,a_{n+1}=\frac{1}{3-a_n}$ satisfies $0<a_n\le 2$ and is decreasing and deduce that the sequence is convergent and find its limit.

Answer to Problem 56E

  $\underset{n\to \infty }{\lim }{a}_n=\frac{3-\sqrt{5}}{2}.$

Explanation of Solution

Given information: The given sequence is $a_1=2,a_{n+1}=\frac{1}{3-a_n}$ .

Calculation:

Let’s assume:

  $\begin{array}{l}0<a_n\le 2\\ 0>-a_n\ge -2\\ -2<-a_n\le 0\\ 1<3-a_n\le 3\\ \frac{1}{3}<\frac{1}{3-a_n}\le 1\\ \frac{1}{3}<a_{n+1}\le 1.\end{array}$

Now since the above is true the following is definitely true:

  $\begin{array}{l}0<a_{n+1}\le 2\\ \text{So}\text{proved}\text{that}{a}_n\in \{0,2\}:\end{array}$

Next step is to prove that $a_n$ is decreasing therefore let’s assume:

  $a_{n+1}<a_n$

Now if prove that the above expression is true for $a_{n+2}<a_{n+1}$ ,it means it is always true:

  $\begin{array}{l}{a}_{n+2}<a_{n+1}\\ \frac{1}{3-a_{n+1}}<\frac{1}{3-a_n}\end{array}$ .

Since

  $\underset{n\to \infty }{\lim }{a}_{n+1}=\frac{1}{3-\underset{n\to \infty }{\lim }{a}_n}.$

Let

  $L=\underset{n\to \infty }{\lim }{a}_{n+1}.$ Then

  $L=\underset{n\to \infty }{\lim }{a}_n$

Plugging in equations:

  $\begin{array}{l}L=\frac{1}{3-L}.\\ L^2-3L+1=0\\ \text{Second}\text{}\text{power}\text{equation}\text{with}a=1,b=-3,c=1:\\ D^2=b^2-4ac=5\\ L_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{3\pm \sqrt{5}}{2}\end{array}$

Reject the bigger value L because it exceeds limitations $a_n\in (0,2)$ .

  $\underset{n\to \infty }{\lim }{a}_n=\frac{3-\sqrt{5}}{2}.$