Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.6 Exercises: 31

Answer

The series is divergent.

Work Step by Step

Use the Ratio Test:$\lim\limits_{n \to \infty}\Big|\frac{a_{n+1}}{a_n}\Big|$. Use the recursive definition $ a_{n+1}=\frac{5n+1}{4n+3}a_n$. Using substitution, $\lim\limits_{n \to \infty}\Big|\frac{\frac{5n+1}{4n+3}a_n}{a_n}\Big|$. The $a_n$ terms cancel, and we are left with $\lim\limits_{n \to \infty}\Big|\frac{5n+1}{4n+3}\Big|=\frac{5}{4}>1$. Since this limit is greater than one, the $\Sigma a_n$ is Divergent by the Ratio Test.
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