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: 19

Answer

\[\text{Converges}\]

Work Step by Step

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