Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 758: 88

Answer

(a) $\frac{1}{f_{n-1}f_{n}}-\frac{1}{f_{n}f_{n+1}}=\frac{1}{f_{n-1}f_{n+1}}$ (b) $\Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}f_{n+1}}=1$ (c) $\Sigma^{\infty}_{n=2} \frac{f_{n}}{f_{n-1}f_{n+1}} = 2$

Work Step by Step

a) $\frac{1}{f_{n-1}f_{n}}-\frac{1}{f_{n}f_{n+1}} =\frac{f_{n+1}}{f_{n-1}f_{n}f_{n+1}}-\frac{f_{n-1}}{f_{n-1}f_{n}f_{n+1}}$ $=\frac{f_{n+1}-f_{n-1}}{f_{n-1}f_{n}f_{n+1}}$ Given $f_{n} = f_{n-1}+f_{n-1}$ $f_{n+1} = f_{n} + f_{n-1}$ $f_{n} = f_{n+1} - f_{n-1}$ Thus, $=\frac{f_{n}}{f_{n-1}f_{n}f_{n+1}}$ $=\frac{1}{f_{n-1}f_{n+1}}$ b) From part (a) we know that $\frac{1}{f_{n-1}f_{n+1}} = \frac{1}{f_{n-1}f_{n}} - \frac{1}{f_{n}f_{n+1}}$ $\Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}f_{n+1}} = \Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}f_{n}} - \frac{1}{f_{n}f_{n+1}}$ $\Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}f_{n}}-\frac{1}{f_{n}f_{n+1}} = \lim\limits_{N \to \infty} \Sigma^{N}_{n=2} \frac{1}{f_{n-1}f_{n}} - \frac{1}{f_{n}f_{n+1}}$ $=\lim\limits_{N \to \infty} \frac{1}{f_{1}f_{2}}- \frac{1}{f_{2}f_{3}} + \frac{1}{f_{2}f_{3}} - \frac{1}{f_{3}f_{4}}+\frac{1}{f_{3}f_{4}} -\frac{1}{f_{4}f_{5}} +...+ \frac{1}{f_{N-1}f_{N}}-\frac{1}{f_{N}f_{N+1}}$ $=\lim\limits_{N \to \infty} \frac{1}{f_{1}f_{2}} - \frac{1}{f_{N}f_{N+1}}$ $=\frac{1}{1 \times 1} - \frac{1}{\infty}$ $ = 1-0 $ $= 1$ c) From part (a) we know that $\frac{1}{f_{n-1}f_{n+1}} = \frac{1}{f_{n-1}f_{n}} - \frac{1}{f_{n}f_{n+1}}$ $\frac{f_{n}}{f_{n-1}f_{n+1}} = \frac{1}{f_{n-1}} - \frac{1}{f_{n+1}}$ Therefore $\Sigma^{\infty}_{n=2} \frac{f_{n}}{f_{n-1}f_{n+1}} = \Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}} - \frac{1}{f_{n+1}}$ $\Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}}-\frac{1}{f_{n+1}} = \lim\limits_{N \to \infty} \Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}}-\frac{1}{f_{n+1}}$ $= \lim\limits_{N \to \infty} \frac{1}{f_{1}}-\frac{1}{f_{3}} + \frac{1}{f_{2}}- \frac{1}{f_{4}}+ \frac{1}{f_{3}}- \frac{1}{f_{5}}+...+ \frac{1}{f_{N-2}} - \frac{1}{f_{N}} + \frac{1}{f_{N-1}}- \frac{1}{f_{N+1}}$ $=\lim\limits_{N \to \infty} \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{1}{f_{N}} - \frac{1}{f_{N+1}}$ $= \frac{1}{1} + \frac{1}{1} - \frac{1}{\infty}- \frac{1}{\infty}$ $=1+1-0-0$ $=2$ Hence, (a) $\frac{1}{f_{n-1}f_{n}}-\frac{1}{f_{n}f_{n+1}}=\frac{1}{f_{n-1}f_{n+1}}$ (b) $\Sigma^{\infty}_{n=2} \frac{1}{f_{n-1}f_{n+1}}=1$ (c) $\Sigma^{\infty}_{n=2} \frac{f_{n}}{f_{n-1}f_{n+1}} = 2$
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