Answer
Converges.
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.
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 1
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