Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 803: 40

Answer

The series has a radius of convergence $5$ and interval $[-5,5]$.

Work Step by Step

Root test:$\lim\limits_{n \to \infty}\sqrt[n] {\frac{|x^{n}|}{n^{2}5^{n}}}=\frac{|x|}{5}\lt 1$ $|x|\lt 5$ $-5\lt x\lt 5$ Radius of convergence = 5 For $x=5$ $\Sigma_{1}^{\infty}(-1)^{n}\frac{5^{n}}{n^{2}5^{n}}=\Sigma_{1}^{\infty}(-1)^{n}\frac{1}{n^{2}}(\frac{5}{5})^{n}$ $=\Sigma_{1}^{\infty}(-1)^{n}\frac{1}{n^{2}}$ Thus, the series has a radius of convergence $5$ and interval $[-5,5]$.
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