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 727: 43

Answer

(a) $s_{n}\leq 1+ln(n)$ (b) $s_{m}\leq 1+ln(10^{6})=1+6 ln 10\approx 14.816\lt 15$ where $m =10^{6}$ = 1 Million and $s_{b}\leq 1+ln(10^{9})=1+9 ln 10\approx 21.723\lt 22$ where $b =10^{9}$ = 1 Billion

Work Step by Step

(a) $\frac{1}{2}+\frac{1}{3}+....+\frac{1}{k}+...+\frac{1}{n}\lt \int_{1}^{n}\frac{dx}{x}$ $\frac{1}{2}+\frac{1}{3}+....+\frac{1}{k}+...+\frac{1}{n}\lt ln(n)$ Add $1$ to both sides. $1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{k}+...+\frac{1}{n}\leq 1+ln(n)$ LHS is $s_{n}$ Thus, $s_{n}\leq 1+ln(n)$ (b) The equality occurs only when $n=1$ Therefore, we have $s_{m}\leq 1+ln(10^{6})=1+6 ln 10\approx 14.816\lt 15$ where $m =10^{6}$ = 1 Million and $s_{b}\leq 1+ln(10^{9})=1+9 ln 10\approx 21.723\lt 22$ where $b =10^{9}$ = 1 Billion
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