## Calculus: Early Transcendentals 8th Edition

Consider $f(x)=\frac{1}{x}$ This function is continuous $[1,\infty)$ and is also decreasing on $[1,\infty)$ (because $f'(x)=\frac{1}{x^{2}}$ , which is always negative for any $x\ne 0$ ). Also $\lim\limits_{n \to \infty}f(x)=0$ However, $\int^{\infty}_{1}f(x)dx =\lim\limits_{x' \to \infty}\int^{x'}_{1}f(x)dx$ $=\lim\limits_{x' \to \infty}\int^{x'}_{1}\frac{1}{x}dx$ $=\lim\limits_{x' \to \infty} [lnx]^{x'}_{1}$ $=\lim\limits_{x' \to \infty} [lnx'-ln1]$ $=\lim\limits_{x' \to \infty} lnx'$ $=\infty$ Therefore, this integral is not convergent. Hence, the given statement is false.