## Calculus: Early Transcendentals 8th Edition

(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
(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