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: 43

Answer

Diverges

Work Step by Step

The $n$-th Term Test states that when a series $a_n \to 0$ then the series diverges. We notice that $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (-1)^n (\sqrt {n+\sqrt n}-\sqrt n) \\=\lim\limits_{n \to \infty} (-1)^n (\sqrt {n+\sqrt n}-\sqrt n) \times \dfrac{(\sqrt {n+\sqrt n}+\sqrt n)}{(\sqrt {n+\sqrt n}+\sqrt n)} \\=\lim\limits_{n \to \infty} (-1)^n \dfrac{\sqrt n}{(\sqrt {n+\sqrt n}+\sqrt n)} \\=\lim\limits_{n \to \infty} (-1)^n (\dfrac{1}{2}) \ne 0$ This implies that the series diverges by the $n$-th Term Test.
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