Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 759: 20

Answer

Convergent

Work Step by Step

Step 1: Restate The Problem Step 2: Create a function f(n) using the series Step 3: Use Divergence Test Problem: $\sum^\infty_{n=1} \frac{1}{n^2+2n+2}$ Test for Divergence Using Divergence Test: $f(n) = \frac{1}{n^2+2n+2}$ Series is convergent if $ \lim\limits_{n \to \infty} f(n) =0$ $\lim\limits_{n \to \infty} \frac{1}{n^2+2n+2} = \frac{1}{\infty} = 0$ Thus, Series is convergent by Divergence Test
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