Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.4 The Comparison Tests - 11.4 Exercises - Page 772: 38


$\Sigma_{n=2}^{\infty}\frac{1}{n^{p} ln(n)}$ converges if $p\gt 1$

Work Step by Step

If $p\lt 0$ then the series converges , since $\lim\limits_{n \to \infty}\frac{1}{n^{p}ln n}=\infty$ if $0\leq p\leq 1$, $n^{p}ln (n)\leq n ln (n)$ and $\Sigma_{n=2}^{\infty}\frac{1}{n ln(n)}$ diverges, so, $\Sigma_{n=2}^{\infty}\frac{1}{n^{p} ln(n)}$ converges if $p\gt 1$
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