## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 557: 45

Diverges

#### Work Step by Step

Given $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+\ln n}$$ Compare with the divergent series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n} }$ ( $p-$ series , p<1) and by using the Limit Comparison Test, we get: \begin{align*} \lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{\sqrt{n}}{\sqrt{n}+\ln n}\ \ \cr & \text{Using l'Hopital's rule}\\ &=\lim_{n\to \infty}\frac{\frac{1}{2 \sqrt{n}}}{\frac{1}{2 \sqrt{n}}+\frac{1}{n}}\\ &=1 \end{align*} Then $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+\ln n}$ also diverges

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