University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 21


$\Sigma_{n=1}^\infty nx^{n-1}=\Sigma_{n=0}^\infty (n+1)x^{n}$

Work Step by Step

Since, we know that the Taylor Series for $\dfrac{1}{1-x}$ is defined as: $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$ Now, $ (\dfrac{1}{1-x})'=[1+x+x^2+....+x^n]'$ or, $\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$ Thus, we have $=\Sigma_{n=1}^\infty nx^{n-1}=\Sigma_{n=0}^\infty (n+1)x^{n}$
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