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 10.6 - Alternating Series and Conditional Convergence - Exercises 10.6 - Page 603: 3

Answer

Converges

Work Step by Step

Let us consider $a_n=\dfrac{1}{ n3^{n}}$and $u_n$ refers to all positive values of $n$. Here, $f(n)=\dfrac{1}{(n)(3^{n})}$; $f'(n)=\dfrac{-3^n(1+\ln 3)}{(n^2)(3^{2n})} \lt 0$ The negative sign simplifies that the sequence $u_n$ is not increasing. Then, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty}\dfrac{1}{(n)(3^{n})}=0$ Hence, the series converges by the Alternating Series Test.
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