University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 40

Answer

Diverges

Work Step by Step

Here, $ s_n=(\sqrt 5 -\sqrt 4)+(\sqrt 6 -\sqrt 5)+.....[\sqrt {n+4} -\sqrt {n+3}]=\sqrt{n+4}-2$ Thus, we have $\lim\limits_{n \to \infty} s_n=\sqrt{n+4}-2=\infty$ We see that $s_n \to \infty$ as $n \to \infty$, so, the given series diverges.
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