Chapter 8.1, Problem 54E

a.

To determine

To show: that $\left\{a_n\right\}$ is increasing and bounded above 3 and $\underset{n\to \infty }{\lim }{a}_n$ is exists by using Monotonic Sequence Theorem.

Answer to Problem 54E

Explanation of Solution

Given information: A sequence $\left\{a_n\right\}$ is given by $a_1=\sqrt{2},a_{n+1}=\sqrt{2+a_n}.$

Calculation:

From the given sequence:

  $a_{n+1}=\sqrt{2+\sqrt{2}}\ge \sqrt{2}=a_n$ .

So, the sequence is increasing. Now use mathematical induction in order to show this sequence is bounded above by 3.

  $n=1:a_1=\sqrt{2}\le 3$ , this is true inequality.

Let hypothesis be that $a_n\le 3$ .

  $\begin{array}{c}{a}_n\le 3\le 7\\ 2+a_n\le 9\\ \sqrt{2+a_n}\le 3\Rightarrow a_{n+1}\le 3\end{array}$

So, for every $n\in N,a_n\le 3$ ,which means this sequence is bounded above by the 3.

This sequence is monotonic and bounded because $\sqrt{2}\le a_n\le 3$ for every $n\ge 1$ so, this sequence is convergent.

Therefore $\underset{n\to \infty }{\lim }{a}_n$ is exists.

b.

To determine

To find: the value of $\underset{n\to \infty }{\lim }{a}_n$ .

Answer to Problem 54E

The limit is 2.

Explanation of Solution

Given information: A sequence $\left\{a_n\right\}$ is given by $a_1=\sqrt{2},a_{n+1}=\sqrt{2+a_n}.$

Calculation:

From part (a) this sequence is convergent, so let M be its limit:

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_{n+1}=\sqrt{2+\underset{n\to \infty }{\lim }{a}_n}.\\ M=\sqrt{2+M}.\\ M^2-M-2=0\\ (M+1)(M-2)=0.\end{array}$

Because this sequence is sequence with positive terms, the conclusion is that the limit is also positive so the limit is M =2.