Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.4 Convergence Tests - Exercises Set 9.4 - Page 629: 7

Answer

(a) The integral test is applicable; the series diverges. (b) The integral test is applicable; the series converges.

Work Step by Step

(a) Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{5x + 2}}$. Notice that the series has positive terms. Since $f$ is decreasing and continuous for all $x \ge 1$, the integral test is applicable. Evaluate: $\mathop \smallint \limits_1^\infty \dfrac{1}{{5x + 2}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{5}\left[ {\ln \left( {5x + 2} \right)} \right]_1^p = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{5}\left[ {\ln \left( {5p + 2} \right) - \ln 7} \right] = \infty $ Since $\mathop \smallint \limits_1^\infty \dfrac{1}{{5x + 2}}{\rm{d}}x$ diverges, by Theorem 9.4.4, the series also diverges. (b) Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{1 + 9{x^2}}}$. Since the series has positive terms and $f$ is decreasing and continuous for all $x \ge 1$, the integral test is applicable. Evaluate $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x$. From Eq. (21) of Section 6.7, we know that $\mathop \smallint \limits_{}^{} \dfrac{1}{{1 + {u^2}}}{\rm{d}}u = {\tan ^{ - 1}}u + C$ Therefore, $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{3}\left( {{{\tan }^{ - 1}}3x} \right)_1^p = \mathop {\lim }\limits_{p \to \infty } \dfrac{1}{3}\left( {{{\tan }^{ - 1}}3p - {{\tan }^{ - 1}}3} \right) = \dfrac{1}{3}\left( {\dfrac{\pi }{2} - {{\tan }^{ - 1}}3} \right)$ $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x = \dfrac{\pi }{6} - \dfrac{1}{3}{\tan ^{ - 1}}3$ Since $\mathop \smallint \limits_1^\infty \dfrac{1}{{1 + 9{x^2}}}{\rm{d}}x$ converges, by Theorem 9.4.4, the series also converges.
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