Chapter 8.1, Problem 47E

a.

To determine

To find: the Fibonacci sequence.

Answer to Problem 47E

Fibonacci sequence is:

  $a_1=1\text{and}{a}_2=1$

  $a_n=a_{n-1}+a_{n-2}$ .

Explanation of Solution

Given information: Suppose that rabbits live forever and that every month each pair.

produces a new pair which becomes productive at age 2 months. If we start with one newborn pair.

Calculation:

Let $a_n$ be the number of rabbit pairs in the nth month .From the given data.

  $a_1=1\text{and}{a}_2=1.$

In the nth month, each pair older than 2 months ( $a_{n-2}$ pairs in all).

Produces a new pair to add to the $a_{n-1}$ existing pairs.

So, the number of rabbit pairs in the nth month is:

  $a_n=a_{n-1}+a_{n-2}$ .

So, Fibonacci sequence is:

  $a_1=1\text{and}{a}_2=1$

  $a_n=a_{n-1}+a_{n-2}$ .

b.

To determine

To show: that $a_{n-1}=1+\frac{1}{a_{n-2}}$ and assuming that { $a_n$ } is convergent, find its limit.

Answer to Problem 47E

The limit of { $a_n$ } is $\frac{1+\sqrt{5}}{2}$ .

Explanation of Solution

Given information: Let $a_n=\frac{f_{n+1}}{f_n}$ . And

Fibonacci sequence is:

  $a_1=1\text{and}{a}_2=1$

  $a_n=a_{n-1}+a_{n-2}$

Calculation:

  $a_n=\frac{f_{n+1}}{f_n}\Rightarrow a_{n-1}=\frac{f_n}{f_{n-1}}=\frac{f_{n-1}+f_{n-2}}{f_{n-1}}=1+\frac{f_{n-2}}{f_{n-1}}=1+\frac{1}{\frac{f_{n-1}}{f_{n-2}}}=1+\frac{1}{a_{n-2}}$

Hence proved.

If exist $\underset{n\to \infty }{\lim }{a}_n$ and if it is M , that means limits of $\underset{n\to \infty }{\lim }{a}_{n-1}$ and $\underset{n\to \infty }{\lim }{a}_{n-2}$ exist too and they are equal to M .

So,

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_{n-1}=\underset{n\to \infty }{\lim }1+\frac{1}{\underset{n\to \infty }{\lim }{a}_{n-2}}.\\ M=1+\frac{1}{M}.\\ M^2-M-1=0.\\ (M-\frac{1+\sqrt{5}}{2})(M-\frac{1-\sqrt{5}}{2})=0\end{array}$

Since M must be positive, the conclusion is $M=\frac{1+\sqrt{5}}{2}$ .

So, the limit of { $a_n$ } is $\frac{1+\sqrt{5}}{2}$ .