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: 25


The series converges for all $x$ in the interval $(-2,4)$.

Work Step by Step

We apply the ratio test $$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ (n+1)(x-3)^{n+1} }{ n(x-3)^{n}}|\\ =|x-3|\lim _{n \rightarrow \infty} \frac{n+1}{n}=|x-3|. $$ Hence, the series $\Sigma_{n=1}^{\infty} n(x-3)^{n}$ converges if and only if $|x-3|\lt1$. That is, the interval of convergence is $(-2,4)$. Now, we check the end points: At $x=-2$, then $\Sigma_{n=1}^{\infty} n(x-3)^{n}=\Sigma_{n=2}^{\infty}(-1)^{n}5 n$, diverges (Alternating Series Test). At $x=4$, then $\Sigma_{n=1}^{\infty} n(x-3)^{n}=\Sigma_{n=1}^{\infty} n$, diverges (by the limit test ). Hence, the series converges for all $x$ in the interval $(-2,4)$.
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