Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 845: 60

Answer

The series diverges.

Work Step by Step

The common ratio of a geometric sequence is equal to the quotient of any term and the term before it: $ \ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$ The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ and a geometric series converges if $|r| \lt 1$. where $r$=common ratio and $a_1$= the first term Now, $r=\dfrac{a_2}{a_1} = \dfrac{12}{9}=\dfrac{4}{3}$ Since $r=|\dfrac{4}{3}| \gt 1$, the infinite geometric series Diverges.
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