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: 28


$$2/3\lt x\lt4/3$$

Work Step by Step

Given $$\sum_{n=0}^{\infty} 27^{n}(x-1)^{3 n+2}$$ Since $a_n =27^{n}(x-1)^{3 n+2}$ and $a_{n+1} =27^{n+1}(x-1)^{3 n+5}$, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{27 (27)^{n}(x-1)^{3} (x-1)^{3 n+2}}{(27)^{n}(x-1)^{3 n+2}}\right|\\ &=\lim _{n \rightarrow \infty}\left|27 (x-1)^{3}\right|\\ & =\lim _{n \rightarrow \infty}\left|27 (x-1)^{3}\right|\\ &=27\left|(x-1)^{3}\right| \lim _{n \rightarrow \infty} 1 \\ &=27\left|(x-1)^{3}\right| \end{aligned} Then the series converges for $$27\left|(x-1)^{3}\right| \lt 1 \ \to \ 2/3\lt x\lt 4/3$$ Now, we check the end points. For $x= 2/3 $ $$\sum_{n=0}^{\infty} 27^{n}(x-1)^{3 n+2} = \frac{1}{9}\sum_{n=1}^{\infty}(-1)^n $$ which is an alternating series and $\lim_{n\to \infty }|a_n| \neq 0 $ (divergent). For $x=4/3 $ $$\sum_{n=0}^{\infty} 27^{n}(x-1)^{3 n+2} = \frac{1}{9}\sum_{n=1}^{\infty}1 $$ which diverges. Hence, the interval of convergence is $$2/3\lt x\lt4/3$$
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