University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 7



Work Step by Step

Consider $f(x)=\dfrac{x}{x^2+4}$ 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 $\int_3^\infty \dfrac{x}{x^2+4}dx= \lim\limits_{a \to \infty} \int_3^a \dfrac{x}{x^2+4}dx$ or, $ \lim\limits_{a \to \infty} [\dfrac{1}{2}\ln (x^2+4)]_3^a=\lim\limits_{a \to \infty} [\dfrac{1}{2} \ln (a^2+4)-\dfrac{1}{2} \ln 13]= \infty$ Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{n}{n^2+4}$ is Divergent.
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