Answer
{$nr^{n}$} converges when $|r| \lt 1$
Work Step by Step
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 $|r| \lt 1$
