# Chapter 11 - Infinite Series - 11.5 The Ratio and Root Tests and Strategies for Choosing Tests - Exercises - Page 568: 46

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.

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