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: 33


$$2-1/e\lt x\lt 2+1/e$$

Work Step by Step

Given $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}$$ Since $a_n = e^{n}(x-2)^{n}$ and $a_{n+1} =e^{n+1}(x-2)^{n+1}$, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{e^{n+1}(x-2)^{n+1}}{e^{n}(x-2)^{n}}\right|\\ &=\lim _{n \rightarrow \infty}|e(x-2)|\\ &=e|x-2| \end{aligned} Then the series converges for $$e|x-2|\lt 1 \ \to \ \ 2-1/e\lt x\lt 2+1/e$$ Now, we check the end points. For $x= 2-1/e $ $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}=\sum_{n=1}^{\infty} (-1)^n $$ which diverges. For $x= 2+1/e$ $$\sum_{n=12}^{\infty} e^{n}(x-2)^{n}=\sum_{n=1}^{\infty}(1)^n $$ which is divergent. Hence, the interval of convergence is $$2-1/e\lt x\lt 2+1/e$$
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