Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 556: 33

Answer

Converges

Work Step by Step

Given $$\sum_{n=1}^{\infty}\frac{(\ln n)^{100}}{n^{1.1}} $$ Since for $n\geq1$ \begin{align*} \frac{(\ln n)^{100}}{n^{1.1}}& \leq \frac{\left(n^{0.0001}\right)^{100}}{n^{1.1}}\\ \frac{(\ln n)^{100}}{n^{1.1}} &\leq \frac{n^{0.01}}{n^{1.1}}\\ \frac{(\ln n)^{100}}{n^{1.1}} &\leq \frac{1}{n^{1.09}} \end{align*} Compare with $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1.09}}$, a convergent $p- $series $ (p>1)$; then $\displaystyle\sum_{n=1}^{\infty}\frac{(\ln n)^{100}}{n^{1.1}} $ also converges
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