Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 25



Work Step by Step

Given $$ a_{n}=\frac{1}{2} 3^{n}-\frac{1}{3} 2^{n} $$ Since $$ a_{n+1}=\frac{1}{2} 3^{n+1}-\frac{1}{3} 2^{n+1}$$ Then \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}&=\lim _{n \rightarrow \infty} \frac{\frac{1}{2} 3^{n+1}-\frac{1}{3} 2^{n+1}}{\frac{1}{2} 3^{n}-\frac{1}{3} 2^{n}}\\ &=\lim _{n \rightarrow \infty} \frac{3^{n+2}-2^{n+2}}{3^{n+1}-2^{n+1}}\\ &=\lim _{n \rightarrow \infty} \frac{3-2\left(\frac{2}{3}\right)^{n+1}}{1-\left(\frac{2}{3}\right)^{n+1}}\\ &=\frac{3-0}{1-0}=3 \end{align*}
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