## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 577: 26

#### Answer

$$14/5\leq x\leq 16/5$$

#### Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}$$ Since $a_n = \dfrac{(-5)^{n}(x-3)^{n}}{n^{2}}$ and $a_{n+1} = \dfrac{(-5)^{n+1}(x-3)^{n+1}}{(n+1)^{2}}$, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{-5(-5)^{n} \frac{(x-3)(x-3)^{n}}{(n+1)^{2}}}{(-5)^{n} \frac{(x-3)^{n}}{n^{2}}}\right|\\ &=\lim _{n \rightarrow \infty}\left|-5(x-3) \cdot \frac{n^{2}}{(n+1)^{2}}\right|\\ & =5|(x-3)| \lim _{n \rightarrow \infty} \frac{1}{\left(1+n^{-2}\right)^{2}}\\ &=5|(x-3)| \end{aligned} Then the series converges for $$5|(x-3)|<1 \ \to \ 14/5 \lt x \lt 16/5$$ Test the end points. For $x= 16/5$ $$\sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}= \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}}$$ This is alternating series, such that $a_{n}$ decreases and $\lim _{n \rightarrow \infty}\left|a_{n}\right|=0$, which converges. For $x= 14/5$ $$\sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}= \sum_{n=1}^{\infty}\frac{1}{n^{2}}$$ This is a $p-$series, ($p=2$) , which converges. Hence, the interval of convergence is $$14/5\leq x\leq 16/5$$

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