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 621: 9


convergent $S=\dfrac{1}{6}$

Work Step by Step

Given $$ \sum_{k=1}^{\infty} \frac{1}{9 k^{2}+3 k-2} $$ Since by using partial fractions \begin{aligned} \sum_{k=1}^{\infty} \frac{1}{9 k^{2}+3 k-2} &=\sum_{k=1}^{\infty} \frac{1}{(3 k+2)(3 k-1)} \\ &=\sum_{k=1}^{\infty} \frac{1}{3}\left(\frac{1}{(3 k-1)}-\frac{1}{(3 k+2)}\right) \\ &=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{1}{9 k-3}-\frac{1}{9 k+6}\right) \\ &=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{9-3}-\frac{1}{9 k+6}\right)+\left(\frac{1}{18-3}-\frac{1}{18+6}\right)+\cdots\right.\\ &\left.=\lim _{n \rightarrow \infty}\left(\frac{1}{6}-\frac{1}{9 n+6}\right)+\cdots+\left(\frac{1}{9 n-3}-\frac{1}{9 n+6}\right)\right] \\ &=\lim _{n \rightarrow \infty}\left(\frac{1}{6}-\frac{1}{9 n+6}\right) \\ &=\frac{1}{6}-0 \\ &=\frac{1}{6} \end{aligned} The series is convergent and the sum is $S=\dfrac{1}{6}$
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