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 577: 1


$ R=2 $ $-2\lt x\lt 2$

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\to \infty }\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim_{n\to \infty }\frac{|x|^{n+1}}{2^{n+1}} \cdot \frac{2^{n}}{|x|^{n}}\\ &=\frac{|x|}{2} \end{aligned} Then the series converges for $$\frac{|x|}{2} \lt 1 \ \to -2\lt x\lt2 $$ The radius of convergence is $ R=2 $. Now, we check the end points For $x= -2 $ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n} =\sum_{n=1}^{\infty} (-1)^n $$ which is a divergent alternating series. For $x=2 $ $$\sum_{n=1}^{\infty} \frac{2^{n}}{3 n}(x+3)^{n}= \sum_{n=1}^{\infty}(1)^n $$ which is divergent by the divergence test. Hence, the interval of convergence is $$-2\lt x\lt 2$$
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