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 10.5 - Absolute Convergence; The Ratio and Root Tests - Exercises 10.5 - Page 598: 61

Answer

Converges

Work Step by Step

Let us consider $a_n=\dfrac{(1)(3) ....(2n-1)}{(4^n)(2^n)n!}$ $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(1)(3) ....(2n-1)(2n+1)}{(4^{n+1})(2^{n+1})(n+1)!}}{\dfrac{(1)(3) ....(2n-1)}{(4^n)(2^n)n!}}|$ $\implies \lim\limits_{n \to \infty}|\dfrac{2n+1}{(4)(2) (n+1)}|=\lim\limits_{n \to \infty}|\dfrac{2+\dfrac{1}{n}}{8+\dfrac{8}{n}}|=\dfrac{2+0}{8+0}=\dfrac{1}{4} \lt 1$ Thus, the series converges by the ratio test.
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