## Calculus: Early Transcendentals 8th Edition

{$nr^{n}$} converges when $|r| \lt 1$
If $|r| \geq$ then $r^{n} → \infty$ as $n → \infty$. So ${nr^{n}} → \infty$ And then ${nr^{n}}$ diverges. If $|r| \lt 1$ then $\lim\limits_{x \to \infty} xr^{x} = \lim\limits_{x \to \infty} \frac{x}{r^{-x}}$ Use L'hospital's rule $\lim\limits_{x \to \infty} \frac{x}{r^{-x}}=\lim\limits_{x \to \infty}\frac{1}{(-\ln r) r^{-x}}$ $=\lim\limits_{x \to \infty} \frac{r^{x}}{-\ln r}$ $=0$ (since $r \lt 1$) (then $r^{x}→0$ as $x→\infty$) So $\lim\limits_{x \to \infty} nr^{n} = 0$ for $|r| \lt 1$ Thus $nr^{n}$ converges when $|r| \lt 1$