## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 15

Divergent

#### Work Step by Step

The integral Test: Let $f(x)=\frac{1}{x\sqrt {lnx}}$ $f(x)$ is continuous, positive, and decreasing on $[2,\infty)$ $\int _{2}^{\infty}f(x)dx=\int _{2}^{\infty}\frac{1}{x\sqrt {lnx}}dx$ $=\int _{2}^{\infty}\frac{1}{\sqrt {lnx}}d(lnx)$ $=2\sqrt {lnx}| _{2}^{\infty}$ $=\infty$ Thus, the series is divergent.

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