Chapter 8, Problem 8RE

To determine

To show: That $\left\{a_n\right\}$ is increasing and $a_n<2$ for all $n$ , and deduce that $\left\{a_n\right\}$ is convergent and find its limit.

Answer to Problem 8RE

The value of limit is $\underset{n\to \infty }{\lim }{a}_n=2$ , and so the sequence $a_n$ is convergent.

Explanation of Solution

Given:

It is given that $a_1=1$ , and $a_{n+1}=\frac{1}{3}\left(a_n+4\right)$ .

Calculation:

Assume that

  $a_{n+1}<a_n<2\dots \left(1\right)$ .

For $n=1$ ,

  $\begin{array}{c}{a}_0<a_1<2\\ a_0+4<a_1+4<2+4\\ \frac{1}{3}\left(a_0+4\right)<\frac{1}{3}\left(a_1+4\right)<\frac{1}{3}\left(2+4\right)\end{array}$

  $a_1<a_2<2$ , since $a_{n+1}=\frac{1}{3}\left(a_n+4\right)$ . So,

  $1<\frac{5}{3}<2$

So, the statement (1) is true for $n=1$ .

Assuming that (1) is true for $n=k$ so

  $\begin{array}{l}{a}_{k+1}<a_k<2\\ a_{k+1}+4<a_k+4<2+4\\ \frac{1}{3}\left(a_{k+1}+4\right)<\frac{1}{3}\left(a_k+4\right)<\frac{1}{3}\left(2+4\right)\end{array}$

  $a_k<a_{k+1}<2$ so it is true for $n=k+1$ so by mathematical induction $a_n<a_{n+1}<2$ is true for all $n$ .

Since $\left\{a_n\right\}$ is an increasing sequence then $a_n<a_{n+1}$

Then let $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(a_{n+1}=L\right)$

Then $L=\frac{1}{3}\left(L+4\right)$

Simplify the equation $L=\frac{1}{3}\left(L+4\right)$ .

  $\begin{array}{c}L-\frac{L}{3}=\frac{4}{3}\\ \frac{2L}{3}=\frac{4}{3}\\ 2L=4\\ L=2\end{array}$

Therefore, the value of limit is $\underset{n\to \infty }{\lim }{a}_n=2$ , and so the sequence $a_n$ is convergent.