Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 29


$$-7/2\leq x\lt-5/2$$

Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n}$$ Since $a_n =\frac{2^{n}}{3 n}(x+3)^{n}$ and $a_{n+1} =\frac{2^{n+1}}{3 n+3}(x+3)^{n+1}$, then \begin{aligned} \rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{2^{n+1}(x+3)^{n+1}}{3(n+1)} \cdot \frac{3 n}{2^{n}(x+3)^{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|2(x+3) \cdot \frac{3 n}{3 n+3}\right| \\ &=\lim _{n \rightarrow \infty}\left|2(x+3) \cdot \frac{1}{1+1 / n}\right|\\ &=|2(x+3)| \end{aligned} Then the series converges for $$|2(x+3)|<1 \ \to -\frac{7}{2} \lt x\lt -\frac{5}{2} $$ Now, we check the end points. For $x= -7/2 $ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n} = \sum_{n=1}^{\infty}(-1)^n \frac{1}{3n} $$ which is an alternating series and $\lim_{n\to \infty }|a_n|= 0 $ (converges). For $x=-5/2$ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n} = \sum_{n=1}^{\infty} \frac{1}{3n} $$ which is a divergent p-series. Hence, the interval of convergence is $$-7/2\leq x\lt-5/2$$
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