Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 638: 23

Answer

\[\text{The series diverges}\]

Work Step by Step

\[\begin{align} & \sum\limits_{k=0}^{\infty }{\frac{1}{\sqrt{k+8}}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=\frac{1}{\sqrt{k+8}},\text{ then let }f\left( x \right)=\frac{1}{\sqrt{x+8}} \\ & \text{ }f\left( x \right)=\frac{1}{\sqrt{x+8}}\text{ is continuous and positive for }x\ge 1, \\ & \text{now we will determine if }f\left( x \right)\text{ is decreasing on }\left( 1,\infty \right)\text{, then } \\ & f\left( x \right)=\frac{1}{\sqrt{x+8}} \\ & f\left( x \right)={{\left( x+8 \right)}^{-1/2}} \\ & \text{Differentiating } \\ & f'\left( x \right)=-\frac{1}{2}{{\left( x+8 \right)}^{-3/2}} \\ & \text{Critical point at }x=-8 \\ & \\ & \text{Intervals: }\left( -\infty ,-8 \right)\text{ and }\left( -8,\infty \right) \\ & \text{The domain of the function }f\left( x \right)\text{ is }\left( -8,\infty \right),\text{ then} \\ & \text{*Test for }\left( -8,\infty \right) \\ & f'\left( -7 \right)=-\frac{1}{2}{{\left( -7+8 \right)}^{-3/2}}=-\frac{1}{2}<0 \\ & \text{Therefore}\text{, the function is decreasing for }\left( 1,\infty \right)\text{ } \\ & \text{The function satisfies the conditions for the integral test}\text{.} \\ & \\ & \int_{1}^{\infty }{\frac{1}{\sqrt{x+8}}}dx \\ & \text{Solving improper integral} \\ & \int_{1}^{\infty }{\frac{1}{\sqrt{x+8}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{{{\left( x+8 \right)}^{-1/2}}}dx \\ & \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{\left( x+8 \right)}^{1/2}} \right]_{1}^{b} \\ & \text{ }=\underset{b\to \infty }{\mathop{\lim }}\,\left[ \sqrt{b+8} \right]-\underset{b\to \infty }{\mathop{\lim }}\,\left[ \sqrt{1+8} \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & \text{ }=\infty -3 \\ & \text{ }=\infty \\ & \text{The integral diverges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series diverges} \\ \end{align}\]
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