Thomas' Calculus 13th Edition

Published by Pearson

Chapter 10: Infinite Sequences and Series - Section 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 1

Converges

Work Step by Step

We will have to compare with $\dfrac{1}{n^2+30} \lt \dfrac{1}{n^2}$ $\dfrac{1}{n^2}$ shows a convergent p-series with p= 2. Hence, by the comparison test, we have $\Sigma_1^{\infty}\dfrac{1}{n^2+30}$ is a convergent series

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