Calculus 8th Edition

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

Chapter 11 - Infinite Sequences and Series - 11.4 The Comparison Tests - 11.4 Exercises - Page 771: 15



Work Step by Step

Using the comparison test, we know the series on the left side is always greater than the series on the right side because the numerator on the left if always larger, and the denominator is always smaller. The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. $\Sigma _{n=1}^{\infty}\frac{4^{n+1}}{3^{n}-2)^{4}} \geq \Sigma _{n=1}^{\infty}\frac{4^{n}}{3^{n}}$ Re-arranging the exponents. $\Sigma _{n=1}^{\infty}\frac{4^{n+1}}{3^{n}-2)^{4}} \geq \Sigma _{n=1}^{\infty}(\frac{4}{3})^{n}$ We find that the series on the right side is a geometric-series with $|r|\gt 1$ , so the series on the right side diverges and therefore, the series on the left side diverges also.
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