Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 9

Answer

Convergent

Work Step by Step

The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ The given series is a p-series with $p= \sqrt 2 \gt 1$ and it is convergent. Hence, the series $\sum_{n=1}^{\infty}\frac{1}{n^{\sqrt 2}}$ is convergent.
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