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 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 25

Answer

Converges

Work Step by Step

Let $u_n=(\dfrac{n}{3n+1})^n$ When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing. $\implies u_n \leq (\dfrac{n}{3n})^n=(\dfrac{1}{3})^n$ Here, the series $\Sigma_{n=1}^\infty (\dfrac{1}{3})^n$ shows a convergent series by the comparison test.
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