Chapter 8.1, Problem 60E

To determine

The first 8 terms of the given sequence and show that the sequence is convergent.

Answer to Problem 60E

The first 8 terms of the given sequence are- $1.5$ , $1.4$ , $1.41667$ , $1.41379$ , $1.41429$ , $1.41420$ and $1.41422$ . The given sequence is convergent and converges at $\sqrt{2}$ .

Explanation of Solution

Given: The given sequence is recursively defined as-

  $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ .

Concepts Used:If $\left\{a_n\right\}$ is convergent sequence then $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{a}_{n+1}$

Calculations: The given sequence is defined as-

  $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ .

Therefore, the first 8 terms of the sequence are-

  $a_1=1$

  $a_2=1+\frac{1}{1+a_1}=1+\frac{1}{1+1}=\frac{3}{2}=1.5$

  $a_3=1+\frac{1}{1+a_2}=1+\frac{1}{1+\frac{3}{2}}=\frac{7}{5}=1.4$

  $a_4=1+\frac{1}{1+a_3}=1+\frac{1}{1+\frac{7}{5}}=\frac{17}{12}=1.41667$

  $a_5=1+\frac{1}{1+a_4}=1+\frac{1}{1+\frac{17}{12}}=\frac{41}{29}=1.41379$

  $a_6=1+\frac{1}{1+a_5}=1+\frac{1}{1+\frac{41}{29}}=\frac{99}{70}=1.41429$

  $a_7=1+\frac{1}{1+a_6}=1+\frac{1}{1+\frac{99}{70}}=\frac{239}{169}=1.41420$

  $a_8=1+\frac{1}{1+a_7}=1+\frac{1}{1+\frac{239}{169}}=\frac{577}{408}=1.41422$

Carefully examining the odd and even terms shows that the given sequence is an alternating sequence. The odd terms are greater in value than the succeeding even terms.

Let,

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

Thus,

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

  $\Rightarrow L=\underset{n\to \infty }{\lim }\left[1+\frac{1}{1+a_n}\right]$

  $\Rightarrow L=\left[1+\frac{1}{1+\underset{n\to \infty }{\lim }{a}_n}\right]$

  $\Rightarrow L=\left[1+\frac{1}{1+L}\right]$ (Since, $\underset{n\to \infty }{\lim }{a}_n=L$ )

  $\Rightarrow L-1=\frac{1}{1+L}$

  $\Rightarrow \left(L-1\right)\left(L+1\right)=1$

  $\Rightarrow L^2-1=1$

  $\Rightarrow L^2=2$

  $\Rightarrow L=\sqrt{2}$

  $\Rightarrow \underset{n\to \infty }{\lim }{a}_n=\sqrt{2}$

Thus, the given sequence is convergent and it converges at $\sqrt{2}$ .