Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 612: 7

Answer

$a.\quad R=1. \quad $Interval of convergence:$\quad -1\lt x\lt 1$ $b.\quad $ Interval of absolute convergence:$\quad -1\lt x\lt 1$ $c.\quad $ No values of x for which the series converges conditionally.

Work Step by Step

(See text: "How to Test a Power Series for Convergence".) $\text{Step 1.}$ Use the Ratio Test to find the interval where the series converges absolutely. Ordinarily, $|x-a|\lt R\quad $ or $\quad a-R\lt x\lt a+R.$ $\begin{align*} \displaystyle \frac{u_{n+1}}{u_{n}}&=\displaystyle \frac{(n+1)x^{n+1}}{(n+3)}\div\frac{nx^{n}}{(n+2)}\\ &=\displaystyle \frac{(n+1)x^{n+1}}{(n+3)}\div\frac{n+2}{nx^{n}}\\ &=\displaystyle \frac{x(n+1)(n+2)}{n(n+3)}\\ \displaystyle \lim_{n\rightarrow\infty}|\frac{x(n+1)(n+2)}{n(n+3)}|&\lt 1\displaystyle \\ |x|\displaystyle \lim_{n\rightarrow\infty}|\frac{(n+1)(n+2)}{n(n+3)}|&\lt 1\displaystyle \\ \end{align*}.$ We see that both polynomials have degree 2. The leading coefficients are the same, so the limit equals 1. $\begin{align*} |x|\cdot 1&\lt 1\\ -1&\lt x\lt 1\\ \end{align*}$ Determine the center and radius: $a-R\lt x\lt a+R$ $0-1\lt x\lt 0+1$ $a=0,\quad $ the radius is $R=1,$ the interval of absolute convergence is $-1\lt x\lt 1$ $\text{Step 2.}$ If the interval of absolute convergence is finite, test for convergence or divergence at each endpoint. $ x=-1\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+2}\qquad$ $\displaystyle \lim_{n\rightarrow\infty}\frac{(-1)^{n}n}{n+2} $ ... is not zero (it fails to exist, alternates to $\pm 1$) ... by the n-th term test, this is a divergent series $ x=1\quad \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+2}\qquad$ ... divergent series (the limit of the n-th term is 1, not 0) $\text{Step 3.}$ If the interval of absolute convergence is $a-R\lt x\lt a+R$, the series diverges for $|x-a|\gt R.$ So, $a.\quad R=1. \quad $Interval of convergence:$\quad -1\lt x\lt 1$ $b.\quad $ Interval of absolute convergence:$\quad -1\lt x\lt 1$ $c.\quad $ No values of x for which the series converges conditionally.
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