Answer
The series $\Sigma_{n=1}^{\infty} 2^{1/n}$ diverges.
Work Step by Step
We have
$$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} 2^{1/n}=2^0=1\ne 0$$
Hence the series $\Sigma_{n=1}^{\infty} 2^{1/n}$ diverges.
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 46
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