Chapter 8.2, Problem 66E

(a)

To determine

To Find: The limit of the sequence.

Answer to Problem 66E

The series increases and it is bounded this shows that the formula $a_n=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)$ for $n\ge 3$ is convergent.

Explanation of Solution

Given:

The sequence $\left\{a_n\right\}$ is defined recursively by the equation $a_n=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)$ for $n\ge 3$ where $a_1$ and $a_2$ , then use the calculator to guess the limit of the sequence.

Calculation:

The given equation is,

  $a_n=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)$

Then,

  $\begin{array}{c}{a}_3=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)\\ =\frac{1}{2}\left(a_2+a_1\right)\\ =\frac{1}{2}\left(1+2\right)\\ =\frac{3}{2}\end{array}$

Then, for $a_4$ is,

  $\begin{array}{c}{a}_4=\frac{1}{2}\left(a_3+a_2\right)\\ =\frac{7}{4}\end{array}$

Then, as the series increases and it is bounded this shows that the formula $a_n=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)$ for $n\ge 3$ is convergent.

(b)

To determine

To Find: The value of $\underset{n\to \infty }{\lim }=a_n$ in terms of the $a_1$ and $a_2$ by the expressing $a_{n+1}-a_n$ in terms of $a_2-a_1$ and then summing the series.

Answer to Problem 66E

The value of the limit is $L=\frac{1}{3}\left(2a_2+a_1\right)$ .

Explanation of Solution

Calculation:

Consider the formula $a_n=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)$ then,

  $\begin{array}{c}{a}_{n-1}-a_n=\frac{1}{2}\left(a_n+a_{n-1}\right)-a_n\\ =-\frac{1}{2}\left(a_n-a_{n-1}\right)\\ :\\ ={\left(-\frac{1}{2}\right)}^{n-1}\left(a_2-a_1\right)\end{array}$

Then,

  $\begin{array}{c}2a_{n+1}+a_n=2a_n+a_{n-1}\\ =2a_{n-1}+a_{n-2}\\ :\\ =2a_2+a_1\end{array}$

Then, consider the limit as $\underset{n\to \infty }{\lim }a=L$ then,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left(2a_{n+1}+a_n\right)=\underset{n\to \infty }{\lim }\left(2a_2+a_1\right)\\ 2L+L=2a_2+a_1\\ L=\frac{1}{3}\left(2a_2+a_1\right)\end{array}$