Chapter 8.2, Problem 64E

(a)

To determine

To Prove: The give statements is true.

Explanation of Solution

Given:

The given statement is,

  $\frac{1}{f_{n-1}{f}_{n+1}}=\frac{1}{f_{n-1}{f}_{n+1}}-\frac{1}{f_n{f}_{n+1}}$

Calculation:

Consider the given Fibonacci sequence is,

  $\frac{1}{f_{n-1}{f}_{n+1}}=\frac{1}{f_{n-1}{f}_{n+1}}-\frac{1}{f_n{f}_{n+1}}$

Then,

  $\begin{array}{c}\frac{1}{f_{n-1}{f}_{n+1}}-\frac{1}{f_n{f}_{n+1}}=\frac{f_{n+1}-f_{n-1}}{f_{n-1}{f}_n{f}_{n+1}}\\ =\frac{f_n+f_{n-1}-f_{n-1}}{f_{n-1}{f}_n{f}_{n+1}}\\ =\frac{f_n}{f_{n+1}{f}_n{f}_{n+1}}\\ =\frac{1}{f_{n+1}{f}_{n+1}}\end{array}$

Hence, Proved.

(b)

To determine

To Prove: The give statements is true.

Explanation of Solution

Given:

The given statement is,

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}{f}_{n+1}}}=1$

Calculation:

Consider the given series is,

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}{f}_{n+1}}}=1$

Then,

  $\begin{array}{l}{\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}{f}_{n+1}}}=\frac{1}{f_1{f}_3}+\frac{1}{f_2{f}_4}+\frac{1}{f_3{f}_5}+\mathrm{....}\frac{1}{f_n{f}_{n+1}}+\mathrm{....}\infty \\ =\underset{n\to \infty }{\lim }\left[\left(\frac{1}{f_1{f}_2}-\frac{1}{f_2{f}_3}\right)+\left(\frac{1}{f_2{f}_3}-\frac{1}{f_3{f}_4}\right)\mathrm{..}+\mathrm{...}++\left(\frac{1}{f_{n-1}{f}_{n+1}}-\frac{1}{f_n{f}_{n+1}}\right)\right]\end{array}$

Then, from part (a),

  $\begin{array}{c}{\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}{f}_{n+1}}}=\underset{n\to \infty }{\lim }\left[\frac{1}{f_1{f}_2}-\left(\frac{1}{f_2{f}_3}-\frac{1}{f_2{f}_3}\right)-\left(\frac{1}{f_3{f}_4}-\frac{1}{f_3{f}_4}\right)-\mathrm{......}-\left(\frac{1}{f_{n-1}{f}_n}-\frac{1}{f_{n-1}{f}_n}\right)-\frac{1}{f_n{f}_{n+1}}\right]\\ =\underset{n\to \infty }{\lim }\left[\frac{1}{f_1{f}_2}-\frac{1}{f_n{f}_{n-1}}\right]\end{array}$

Then, by the Fibonacci series,

  $\begin{array}{c}{\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}{f}_{n+1}}}=\left[\frac{1}{f_1{f}_2}-\frac{1}{\underset{n\to \infty }{\lim }{f}_n{f}_{n-1}}\right]\\ =\frac{1}{1\cdot 1}-0\\ =1\end{array}$

Hence, proved.

(c)

To determine

To Prove: The give statements is true.

Explanation of Solution

Given:

The given statement is,

  ${\displaystyle \sum _{n=2}^{\infty }\frac{f_n}{f_{n-1}{f}_{n+1}}}=2$

Calculation:

Consider the given equation from part (1) is,

  $\frac{1}{f_{n-1}{f}_{n+1}}-\frac{1}{f_n{f}_{n+1}}=\frac{1}{f_{n+1}{f}_{n+1}}$

Then,

  $\begin{array}{c}\frac{1}{f_{n-1}{f}_{n+1}}=\frac{1}{f_{n+1}{f}_{n+1}}-\frac{f_n}{f_n{f}_{n+1}}\\ =\frac{1}{f_{n-1}}-\frac{1}{f_{n+1}}\end{array}$

Then,

  $\begin{array}{c}{\displaystyle \sum _{n=2}^{\infty }\frac{f_n}{f_{n-1}{f}_{n+1}}}={\displaystyle \sum _{n=2}^{\infty }\frac{1}{f_{n-1}}-\frac{1}{f_{n+1}}}\\ =\underset{n\to \infty }{\lim }\left[\left(\frac{1}{f_1}-\frac{1}{f_3}\right)+\left(\frac{1}{f_2}-\frac{1}{f_4}\right)+\left(\frac{1}{f_3}-\frac{1}{f_5}\right)\mathrm{.....}+\left(\frac{1}{f_{n-1}}-\frac{1}{f_{n+1}}\right)\right]\\ =\underset{n\to \infty }{\lim }\left[\frac{1}{f_1}+\frac{1}{f_2}-\frac{1}{f_n}-\frac{1}{f_{n+1}}\right]\end{array}$

Consider that by Fibonacci series $\underset{n\to \infty }{\lim }{f}_n=\infty$ , then $\underset{n\to \infty }{\lim }\frac{1}{f_{n+1}}=0$ that is $f_1=f_2$ which is equal to 0. Then,

  $\begin{array}{c}{\displaystyle \sum _{n=2}^{\infty }\frac{f_n}{f_{n-1}{f}_{n+1}}}=\underset{n\to \infty }{\lim }\left[\frac{1}{f_1}+\frac{1}{f_2}-\frac{1}{f_n}-\frac{1}{f_{n+1}}\right]\\ =\frac{1}{1}+\frac{1}{1}-0-0\\ =2\end{array}$

Hence, proved.