Chapter 8.1, Problem 55E

To determine

Show that: The sequence is increasing and $a<3$ for all n.

Deduce that: The sequence $\left\{a_n\right\}$ is convergent and find its limit.

Answer to Problem 55E

The limit of the sequence is $1$ .

Explanation of Solution

Given information: The sequence defined by $a_1=1\text{ and }{a}_{n+1}=3-\frac{1}{a_n}$

Definitions:

Monotone increasing: A sequence ${\left\{x_n\right\}}_n$ is said to be monotone increasing, iff $\forall n,x_{n+1}\ge x_n.$

Monotone decreasing: A sequence ${\left\{x_n\right\}}_n$ is said to be monotone decreasing iff $\forall n,x_{n+1}\le x_n.$

Prove: Given sequence $a_{n+1}=3-\frac{1}{a_n}$

Also given $a_1=1$

Now put $n=1$ in the given sequence $a_{n+1}=3-\frac{1}{a_n}$

So,

  $\begin{array}{l}\Rightarrow a_{1+1}=3-\frac{1}{a_1}\\ \Rightarrow a_2=3-\frac{1}{1}\\ \Rightarrow a_2=3-1\\ \Rightarrow a_2=2\end{array}$

From the above definition, we have $a_2-a_1$

  $\begin{array}{l}\Rightarrow 2-1\\ \Rightarrow 1\end{array}$

Thus $1>0$

From the above definition , the sequence is monotone increasing .

Again, let $3-\frac{1}{a_n}\le 2$

  $\begin{array}{l}\Rightarrow 3-\frac{1}{a_n}\le 2\\ \Rightarrow 3a_n-1\le 2a_n\\ \Rightarrow 3a_n-2a_n\le 1\\ \Rightarrow a_n\le 1\end{array}$

So, form the above inequality its prove that $a_n<3$ .

From the above solution the sequence is monotone increasing and bounded above.

So, the sequence $\left\{a_n\right\}$ is convergent.

From the above inequality , the limit of the sequence is $1$ .