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 577: 22


The radius of converges is $R=1$ and converges for $-1\leq x\lt 1$.

Work Step by Step

Apply the ratio test: \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ & =\lim _{n \rightarrow \infty}\left|\frac{\frac{x \times x^{n}}{(n+1)-4 \ln (n+1)}}{\frac{x^{n}}{n-4 \ln n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|x \cdot \frac{n-4 \ln n}{(n+1)-4 \ln (n+1)}\right|\\ & =|x| \times \lim _{n \rightarrow \infty} \frac{n-4 \ln n}{(n+1)-4 \ln (n+1)}\\ &= |x| \times \lim _{n \rightarrow \infty} \frac{1-\frac{4}{n}}{1-\frac{4}{n+1}} \end{aligned} Then the radius of converges is $R=1$ and the series converges for $-1\lt x\lt 1$. For $x=1 $, the series $\sum_{n=9}^{\infty} \frac{1}{n-4 \ln n} $ diverges by using the comparison test with $\sum_{n=9}^{\infty} \frac{1}{n } $. For $x=-1 $, the series $\sum_{n=9}^{\infty} \frac{(-1)^n}{n-4 \ln n} $ is an alternating series; when n increases, the value of $a_{n}$ decreases and $\lim _{n \rightarrow \infty}\left|a_{n}\right|=0$. So, the using Alternating Series Test, we can say that the series converges.
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