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: 21

Answer

\[\text{The series converges}\]

Work Step by Step

\[\begin{align} & \sum\limits_{k=1}^{\infty }{k{{e}^{-2{{k}^{2}}}}} \\ & \sum\limits_{k=1}^{\infty }{{{a}_{k}}\Rightarrow }\text{ }{{a}_{k}}=k{{e}^{-2{{k}^{2}}}},\text{ then let }f\left( x \right)=x{{e}^{-2{{x}^{2}}}} \\ & \text{We know that }f\left( x \right)=x{{e}^{-2{{x}^{2}}}}\text{ is continuous and positive for }x\ge 1, \\ & \text{now we will determine if }f\left( x \right)\text{ is decreasing}\text{, then } \\ & f\left( x \right)=x{{e}^{-2{{x}^{2}}}} \\ & \text{Differentiating by the product rule} \\ & f'\left( x \right)=x\left( -4x{{e}^{-2{{x}^{2}}}} \right)+{{e}^{-2{{x}^{2}}}} \\ & f'\left( x \right)={{e}^{-2{{x}^{2}}}}\left( 1-4{{x}^{2}} \right) \\ & f'\left( x \right)=0 \\ & 1-4{{x}^{2}}=0\to {{x}_{1}}=-\frac{1}{2},\text{ }x=\frac{1}{2} \\ & \text{Test }f'\left( x \right)\text{ for the interval }\left( 1,\infty \right): \\ & f'\left( 2 \right)={{e}^{-2{{\left( 2 \right)}^{2}}}}\left( 1-4{{\left( 2 \right)}^{2}} \right)<0,\text{ decreasing}\text{.} \\ & \text{The function satisfies the conditions for the integral test}\text{.} \\ & \\ & \int_{1}^{\infty }{x{{e}^{-2{{x}^{2}}}}}dx \\ & \text{Solving improper integral} \\ & \int_{1}^{\infty }{x{{e}^{-2{{x}^{2}}}}}dx=\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{x{{e}^{-2{{x}^{2}}}}}dx \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\int_{1}^{b}{{{e}^{-2{{x}^{2}}}}\left( -4x \right)}dx \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{x}^{2}}}} \right]_{1}^{b} \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{\left( b \right)}^{2}}}}-{{e}^{-2}} \right] \\ & \text{Evaluate the limit when }b\to \infty \\ & \text{ }=-\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2{{\left( b \right)}^{2}}}} \right]+\frac{1}{4}\underset{b\to \infty }{\mathop{\lim }}\,\left[ {{e}^{-2}} \right] \\ & \text{ }=0+\frac{1}{4{{e}^{2}}} \\ & \text{ }=\frac{1}{4{{e}^{2}}} \\ & \text{The integral converges}\text{, so by the Integral test }\left( \text{Theorem 8}\text{.10} \right) \\ & \text{The series converges} \\ \end{align}\]
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