Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.4 - The Comparison Tests - 11.4 Exercises - Page 731: 11



Work Step by Step

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 $p\leq 1$. Using the Comparison Test, the series is less than $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}} $ because the denominator of the original series is always slightly larger than the series on the right. Therefore, $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}}} $ Re-arranging the exponents: $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{k^{1/3}}{k^{3/2}} $ $\Sigma _{k=1}^{\infty}\frac{\sqrt[3] k}{\sqrt{ k^{3}+4k+3}} \leq \Sigma _{k=1}^{\infty}\frac{1}{k^{1/6}} $ We find that the series on the right side is a p-series with $p \gt 1$, so the series on the right side converges and therefore, the series on the left converges.
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