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: 53



Work Step by Step

As we are given that $a_{n+1}=\sqrt [n] a_n$ and $a_1=\dfrac{1}{3}$ ; and $n=1,2,3,....n$ Now, we can see that $a_2=\sqrt [1] a_1=(\dfrac{1}{3})^{1/1} \\ a_3=\sqrt [2] {\dfrac{1}{3}}=(\dfrac{1}{3})^{1/2}.... \\ ...a_n=(\dfrac{1}{3})^{(1/n)!} $ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{3})^{(1/n!)}=1$ Thus, the series diverges with limit $1$.
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