## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 6

#### Answer

Converges to $\dfrac{1}{\ln 2}$

#### Work Step by Step

Consider $f(x)=\dfrac{1}{x(\ln x)^2}$. This is positive, continuous and decreasing for $x \geq 1$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\int_2^\infty \dfrac{1}{x(\ln x)^2} dx= \lim\limits_{a \to \infty} \int_2^a \dfrac{1}{x(\ln x)^2} dx$ or, $\lim\limits_{a \to \infty} [\dfrac{-1}{\ln x}]_2^a=\lim\limits_{a \to \infty} [\dfrac{-1}{\ln a} +\dfrac{1}{\ln 2}]= \dfrac{1}{\ln 2}$ Hence, the sequence $\Sigma_{n=2}^\infty \dfrac{1}{n(\ln n)^2}$ converges to $\dfrac{1}{\ln 2}$

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