University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 1



Work Step by Step

We compare with: $\dfrac{1}{n^2+30} \lt \dfrac{1}{n^2}$ We know that $\dfrac{1}{n^2}$ is a convergent p-series with $p=2$ Thus, by the comparison test, $\Sigma_1^\infty \frac{1}{n^2+30}$ converges.
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