Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises - Page 639: 52

Answer

Diverges

Work Step by Step

Recall the property that the series $\Sigma_{n=1}^\infty a_n$ diverges if $\lim\limits_{n \to \infty} a_n \ne 0$. Now, we have : $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \sqrt {\dfrac{k+1}{k}}$ or, $\lim\limits_{k \to \infty} \sqrt {1+\dfrac{1}{k}}=1$ So, we can see that $\lim\limits_{k \to \infty} a_k\ne 0$. Thus, the given series diverges.
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