Chapter 8.1, Problem 46E

a.

To determine

To find: the value of $\underset{n\to \infty }{\lim }{a}_{n+1}$ .

Answer to Problem 46E

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

Explanation of Solution

Given information: Given $\underset{n\to \infty }{\lim }{a}_n=L$ .

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_{n+1}=\underset{m+1\to \infty }{\lim }{a}_m=\underset{m\to \infty }{\lim }{a}_m=\underset{n\to \infty }{\lim }{a}_n=L(\text{Given})\\ \text{Therefore},\underset{n\to \infty }{\lim }{a}_{n+1}=L.\end{array}$

b.

To determine

To find: the first ten term of the sequence correct to five decimal places and does it appears that the sequence is convergent and estimate the value of the limit to three decimal places.

Answer to Problem 46E

  $L=\mathrm{0.618.}$

Explanation of Solution

Given information: The given sequence is defined by:

  $a_1=1,a_{n+1}=\frac{1}{(1+a_n)}\text{for}n\ge 1.$

Calculation:

So, to find the first 10 terms, use the given equation of the sequence:

  $\begin{array}{c}{a}_1=\mathrm{1.00000.}\\ a_2=\frac{1}{1+a_1}=\frac{1}{1+1.00000}=\frac{1}{2}=\mathrm{0.50000.}\\ a_3=\frac{1}{1+a_2}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}=\mathrm{0.66667.}\\ a_4=\frac{1}{1+a_3}=\frac{1}{1+\frac{2}{3}}=\frac{3}{5}=\mathrm{0.60000.}\\ a_5=\frac{1}{1+a_4}=\frac{1}{1+\frac{3}{5}}=\frac{5}{8}=\mathrm{0.62500.}\\ a_6=\frac{1}{1+a_5}=\frac{1}{1+\frac{5}{8}}=\frac{8}{13}=\mathrm{0.61538.}\\ a_7=\frac{1}{1+a_6}=\frac{1}{1+\frac{8}{13}}=\frac{13}{21}=\mathrm{0.61905.}\\ a_8=\frac{1}{1+a_7}=\frac{1}{1+\frac{13}{21}}=\frac{21}{34}=\mathrm{0.61765.}\\ a_9=\frac{1}{1+a_8}=\frac{1}{1+\frac{21}{34}}=\frac{34}{55}=\mathrm{0.61818.}\\ a_{10}=\frac{1}{1+a_9}=\frac{1}{1+\frac{34}{55}}=\frac{55}{89}=\mathrm{0.61798.}\end{array}$

The series does appears convergent and by looking at the last term it appears to converge around the value:

  $L=\mathrm{0.618.}$

c.

To determine

To find: the value of $\underset{n\to \infty }{\lim }{a}_n$ using part (a) and compare this value from estimated value from part (b).

Answer to Problem 46E

  $\underset{n\to \infty }{\lim }{a}_n=0.618$ .

Explanation of Solution

Given information: Given $\underset{n\to \infty }{\lim }{a}_n=L$ .

The given sequence is defined by:

  $a_1=1,a_{n+1}=\frac{1}{(1+a_n)}\text{for}n\ge 1.$

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_{n+1}=\underset{m+1\to \infty }{\lim }{a}_m=\underset{m\to \infty }{\lim }{a}_m=\underset{n\to \infty }{\lim }{a}_n=L(\text{Given})\\ \text{Therefore},\underset{n\to \infty }{\lim }{a}_{n+1}=L.\end{array}$

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_{n+1}=\underset{n\to \infty }{\lim }\frac{1}{1+a_n}.\\ \underset{n\to \infty }{\lim }{a}_n=\frac{1}{1+\underset{n\to \infty }{\lim }{a}_n}.\\ \underset{n\to \infty }{\lim }{a}_n(1+\underset{n\to \infty }{\lim }{a}_n)=1.\\ \text{Let}x=\underset{n\to \infty }{\lim }{a}_n.\\ x(1+x)=1\\ x^2+x-1=0\\ solving,get:\\ x_1=-1.618,\\ x_2=\mathrm{0.618.}\end{array}$

However because n is never negative x cannot be negative so:

  $\begin{array}{l}x=\mathrm{0.618.}\\ \underset{n\to \infty }{\lim }{a}_n=\mathrm{0.618.}\end{array}$

The above is exactly what found in part (b).