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: 27


$$ 6\lt x\lt 8$$

Work Step by Step

Given $$\sum_{n=1}^{\infty}(-1)^{n} n^{5}(x-7)^{n}$$ Since $a_n = (-1)^{n} n^{5}(x-7)^{n}$ and $a_{n+1} = (-1)^{n+1} (n+1)^{5}(x-7)^{n+1}$, then \begin{aligned} \rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1}(n+1)^{5}(x-7)^{n+1}}{(-1)^{n} n^{5}(x-7)^{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|(x-7) \cdot \frac{(n+1)^{5}}{n^{5}}\right| \\ &=\lim _{n \rightarrow \infty}\left|(x-7) \cdot \frac{n^{5}+\ldots}{n^{5}}\right|\\ &=|x-7| \end{aligned} Then the series converges for $$ |x-7|\lt 1 \ \to \ 6 \leq x\leq 8$$ Now, we check the end points. For $x= 6 $ $$\sum_{n=1}^{\infty}(-1)^{n} n^{5}(x-7)^{n}= \sum_{n=1}^{\infty}n^5 $$ which diverges by divergence test For $x= 8 $ $$\sum_{n=1}^{\infty}(-1)^{n} n^{5}(x-7)^{n}= \sum_{n=1}^{\infty}(-1)^nn^5 $$ which is an alternating series and $\lim_{n\to \infty }|a_n| \neq 0 $. Thus, it diverges. Hence, the interval of convergence is $$ 6\lt x\lt 8$$
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