Calculus, 10th Edition (Anton)

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

Chapter 9 - Infinite Series - 9.3 Infinite Series - Exercises Set 9.3 - Page 622: 31

Answer

(a) The series converges if $ - 1 \lt x \lt 1$; $S = \dfrac{x}{{1 + {x^2}}}$. (b) The series converges if $\left| x \right| \gt 2$; $S = \dfrac{1}{{{x^2} - 2x}}$. (c) The series converges if $x \gt 0$; $S = \dfrac{1}{{{{\rm{e}}^x} - 1}}$.

Work Step by Step

(a) Write $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = x - {x^3} + {x^5} - {x^7} + {x^9} - \cdot\cdot\cdot$ $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k}x = x\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k}$ According to Exercise 30 (c): $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {{x^2}} \right)^k} = \dfrac{1}{{1 + {x^2}}}$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$ Therefore $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = x\left( {\dfrac{1}{{1 + {x^2}}}} \right)$ ${\ \ \ }$ if $ - 1 \lt x \lt 1$ Thus, the series converges if $ - 1 \lt x \lt 1$ and its sum is $\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{x^{2k + 1}} = \dfrac{x}{{1 + {x^2}}}$. (b) Write $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2}}} + \dfrac{2}{{{x^3}}} + \dfrac{4}{{{x^4}}} + \dfrac{8}{{{x^5}}} + \dfrac{{16}}{{{x^6}}} + \cdot\cdot\cdot$ $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{x^2}}}{\left( {\dfrac{2}{x}} \right)^k}$ The series on the right-hand side converges when it is a geometric series with $a = \dfrac{1}{{{x^2}}}$ and $\left| r \right| = \left| {\dfrac{2}{x}} \right| \lt 1$. That is, $\left| x \right| \gt 2$ And its sum by 9.3.3 Theorem is $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{x^2}}}{\left( {\dfrac{2}{x}} \right)^k} = \dfrac{{\dfrac{1}{{{x^2}}}}}{{1 - \dfrac{2}{x}}} = \dfrac{1}{{{x^2} - 2x}}$ Therefore, $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2} - 2x}}$ ${\ \ \ }$ if $\left| x \right| \gt 2$ Thus, the series converges if $\left| x \right| \gt 2$ and its sum is $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{2^k}}}{{{x^{k + 2}}}} = \dfrac{1}{{{x^2} - 2x}}$. (c) Write $\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = {{\rm{e}}^{ - x}} + {{\rm{e}}^{ - 2x}} + {{\rm{e}}^{ - 3x}} + {{\rm{e}}^{ - 4x}} + {{\rm{e}}^{ - 5x}} + \cdot\cdot\cdot$ $\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^{\left( {k + 1} \right)x}}}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^x}}}{\left( {\dfrac{1}{{{{\rm{e}}^x}}}} \right)^k}$ The series on the right-hand side converges when it is a geometric series with $a = \dfrac{1}{{{{\rm{e}}^x}}}$ and $\left| r \right| = \left| {\dfrac{1}{{{{\rm{e}}^x}}}} \right| \lt 1$. That is, ${{\rm{e}}^x} \gt 1$ $\ln {{\rm{e}}^x} \gt \ln 1 = 0$ $x \gt 0$ And the sum by 9.3.3 Theorem is $\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{{{{\rm{e}}^x}}}{\left( {\dfrac{1}{{{{\rm{e}}^x}}}} \right)^k} = \dfrac{{\dfrac{1}{{{{\rm{e}}^x}}}}}{{1 - \dfrac{1}{{{{\rm{e}}^x}}}}} = \dfrac{1}{{{{\rm{e}}^x} - 1}}$ Thus, the series converges if $x \gt 0$ and its sum is $\mathop \sum \limits_{k = 0}^\infty {{\rm{e}}^{ - \left( {k + 1} \right)x}} = \dfrac{1}{{{{\rm{e}}^x} - 1}}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.