Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 44


The series $ \Sigma_{n=1}^{\infty} \frac{1}{\sqrt n+n}$ diverges.

Work Step by Step

Take $a_n= \frac{1}{\sqrt n+n}$ and $b_n= \frac{1}{n}$; now we have the limit comparison test $$L=\lim_{n \to \infty}\frac{a_n}{ b_n}=\lim_{n \to \infty}\frac{1/(\sqrt n+n)}{1/n}\\ =\lim_{n \to \infty}\frac{n}{\sqrt n+n}=\lim_{n \to \infty}\frac{1}{(1/\sqrt n)+1}=1$$ Since $L\gt 0$ and the series $ \Sigma_{n=1}^{\infty} \frac{1}{n}$ diverges then the series $ \Sigma_{n=1}^{\infty} \frac{1}{\sqrt n+n}$ diverges.
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