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: 3


converges to $\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$

Work Step by Step

Consider $f(x)=\dfrac{1}{x^2+4}$. 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 $\int_1^\infty \dfrac{1}{x^2+4} dx= \lim\limits_{a \to \infty} \int_1^a \dfrac{1}{x^2+4} dx$ or, $ \lim\limits_{a \to \infty} [\frac{1}{2}\tan^{-1} \dfrac{x}{2}]_1^a=\lim\limits_{a \to \infty} [\frac{1}{2}\tan^{-1} \dfrac{a}{2}-\dfrac{1}{2}\tan^{-1} \dfrac{1}{2}]=\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$ Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}+4}$ converges to $\dfrac{\pi}{4}-\dfrac{1}{2}\tan^{-1}( \dfrac{1}{2})$
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